#include <bits/stdc++.h>
#include <sys/time.h>
using namespace std;

#define rep(i,n) for(long long i = 0; i < (long long)(n); i++)
#define repi(i,a,b) for(long long i = (long long)(a); i < (long long)(b); i++)
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define mt make_tuple
#define mp make_pair
template<class T1, class T2> bool chmin(T1 &a, T2 b) { return b < a && (a = b, true); }
template<class T1, class T2> bool chmax(T1 &a, T2 b) { return a < b && (a = b, true); }
#define exists find_if
#define forall all_of

using ll = long long; using vll = vector<ll>; using vvll = vector<vll>; using P = pair<ll, ll>;
using ld = long double;  using vld = vector<ld>; 
using vi = vector<int>; using vvi = vector<vi>; vll conv(vi& v) { vll r(v.size()); rep(i, v.size()) r[i] = v[i]; return r; }
using Pos = complex<double>;

template <typename T, typename U> ostream &operator<<(ostream &o, const pair<T, U> &v) {  o << "(" << v.first << ", " << v.second << ")"; return o; }
template<size_t...> struct seq{}; template<size_t N, size_t... Is> struct gen_seq : gen_seq<N-1, N-1, Is...>{}; template<size_t... Is> struct gen_seq<0, Is...> : seq<Is...>{};
template<class Ch, class Tr, class Tuple, size_t... Is>
void print_tuple(basic_ostream<Ch,Tr>& os, Tuple const& t, seq<Is...>){ using s = int[]; (void)s{0, (void(os << (Is == 0? "" : ", ") << get<Is>(t)), 0)...}; }
template<class Ch, class Tr, class... Args> 
auto operator<<(basic_ostream<Ch, Tr>& os, tuple<Args...> const& t) -> basic_ostream<Ch, Tr>& { os << "("; print_tuple(os, t, gen_seq<sizeof...(Args)>()); return os << ")"; }
ostream &operator<<(ostream &o, const vvll &v) { rep(i, v.size()) { rep(j, v[i].size()) o << v[i][j] << " "; o << endl; } return o; }
template <typename T> ostream &operator<<(ostream &o, const vector<T> &v) { o << '['; rep(i, v.size()) o << v[i] << (i != v.size()-1 ? ", " : ""); o << "]";  return o; }
template <typename T>  ostream &operator<<(ostream &o, const set<T> &m) { o << '['; for (auto it = m.begin(); it != m.end(); it++) o << *it << (next(it) != m.end() ? ", " : ""); o << "]";  return o; }
template <typename T, typename U>  ostream &operator<<(ostream &o, const map<T, U> &m) { o << '['; for (auto it = m.begin(); it != m.end(); it++) o << *it << (next(it) != m.end() ? ", " : ""); o << "]";  return o; }
template <typename T, typename U, typename V>  ostream &operator<<(ostream &o, const unordered_map<T, U, V> &m) { o << '['; for (auto it = m.begin(); it != m.end(); it++) o << *it; o << "]";  return o; }
vector<int> range(const int x, const int y) { vector<int> v(y - x + 1); iota(v.begin(), v.end(), x); return v; }
template <typename T> istream& operator>>(istream& i, vector<T>& o) { rep(j, o.size()) i >> o[j]; return i;}
string bits_to_string(ll input, ll n=64) { string s; rep(i, n) s += '0' + !!(input & (1ll << i)); reverse(all(s)); return s; }

template <typename T> unordered_map<T, ll> counter(vector<T> vec){unordered_map<T, ll> ret; for (auto&& x : vec) ret[x]++; return ret;};
string substr(string s, P x) {return s.substr(x.fi, x.se - x.fi); }
struct ci : public iterator<forward_iterator_tag, ll> { ll n; ci(const ll n) : n(n) { } bool operator==(const ci& x) { return n == x.n; } bool operator!=(const ci& x) { return !(*this == x); } ci &operator++() { n++; return *this; } ll operator*() const { return n; } };

size_t random_seed; namespace std { using argument_type = P; template<> struct hash<argument_type> { size_t operator()(argument_type const& x) const { size_t seed = random_seed; seed ^= hash<ll>{}(x.fi); seed ^= (hash<ll>{}(x.se) << 1); return seed; } }; }; // hash for various class
namespace myhash{ const int Bsizes[]={3,9,13,17,21,25,29,33,37,41,45,49,53,57,61,65,69,73,77,81}; const int xor_nums[]={0x100007d1,0x5ff049c9,0x14560859,0x07087fef,0x3e277d49,0x4dba1f17,0x709c5988,0x05904258,0x1aa71872,0x238819b3,0x7b002bb7,0x1cf91302,0x0012290a,0x1083576b,0x76473e49,0x3d86295b,0x20536814,0x08634f4d,0x115405e8,0x0e6359f2}; const int hash_key=xor_nums[rand()%20]; const int mod_key=xor_nums[rand()%20]; template <typename T> struct myhash{ std::size_t operator()(const T& val) const { return (hash<T>{}(val)%mod_key)^hash_key; } }; };
template <typename T> class uset:public std::unordered_set<T,myhash::myhash<T>> { using SET=std::unordered_set<T,myhash::myhash<T>>; public: uset():SET(){SET::rehash(myhash::Bsizes[rand()%20]);} };
template <typename T,typename U> class umap:public std::unordered_map<T,U,myhash::myhash<T>> { public: using MAP=std::unordered_map<T,U,myhash::myhash<T>>; umap():MAP(){MAP::rehash(myhash::Bsizes[rand()%20]);} };    

struct timeval start; double sec() { struct timeval tv; gettimeofday(&tv, NULL); return (tv.tv_sec - start.tv_sec) + (tv.tv_usec - start.tv_usec) * 1e-6; }
struct init_{init_(){ gettimeofday(&start, NULL); ios::sync_with_stdio(false); cin.tie(0); srand((unsigned int)time(NULL)); random_seed = RAND_MAX / 2 + rand() / 2; }} init__;

static const double EPS = 1e-14;
static const long long INF = 1e18;
#define ldout fixed << setprecision(40) 

static const long long mo = 1e9+7;
class Mod {
    public:
        long long num;
        Mod() : Mod(0) {}
        Mod(long long int n) : num(n) { }
        Mod(const string &s){ long long int tmp = 0; for(auto &c:s) tmp = (c-'0'+tmp*10) % mo; num = tmp; }
        Mod(int n) : Mod(static_cast<long long int>(n)) {}
        operator int() { return num; }
};
istream &operator>>(istream &is, Mod &x) { long long int n; is >> n; x = n; return is; }
ostream &operator<<(ostream &o, const Mod &x) { o << x.num; return o; }
Mod operator+(const Mod a, const Mod b) { return Mod((a.num + b.num) % mo); }
Mod operator+(const long long int a, const Mod b) { return Mod(a) + b; }
Mod operator+(const Mod a, const long long int b) { return b + a; }
Mod operator++(Mod &a) { return a + Mod(1); }
Mod operator-(const Mod a, const Mod b) { return Mod((mo + a.num - b.num) % mo); }
Mod operator-(const long long int a, const Mod b) { return Mod(a) - b; }
Mod operator--(Mod &a) { return a - Mod(1); }
Mod operator*(const Mod a, const Mod b) { return Mod(((long long)a.num * b.num) % mo); }
Mod operator*(const long long int a, const Mod b) { return Mod(a)*b; }
Mod operator*(const Mod a, const long long int b) { return Mod(b)*a; }
Mod operator*(const Mod a, const int b) { return Mod(b)*a; }
Mod operator+=(Mod &a, const Mod b) { return a = a + b; }
Mod operator+=(long long int &a, const Mod b) { return a = a + b; }
Mod operator-=(Mod &a, const Mod b) { return a = a - b; }
Mod operator-=(long long int &a, const Mod b) { return a = a - b; }
Mod operator*=(Mod &a, const Mod b) { return a = a * b; }
Mod operator*=(long long int &a, const Mod b) { return a = a * b; }
Mod operator*=(Mod& a, const long long int &b) { return a = a * b; }
Mod factorial(const long long n) {
    if (n < 0) return 0;
    Mod ret = 1;
    for (int i = 1; i <= n; i++) {
        ret *= i;
    }
    return ret;
}
Mod operator^(const Mod a, const long long n) {
    if (n == 0) return Mod(1);
    Mod res = (a * a) ^ (n / 2);
    if (n % 2) res = res * a;
    return res;
}
Mod modpowsum(const Mod a, const long long b) {
    if (b == 0) return 0;
    if (b % 2 == 1) return modpowsum(a, b - 1) * a + Mod(1);
    Mod result = modpowsum(a, b / 2);
    return result * (a ^ (b / 2)) + result;
}


/*************************************/
// 以下、modは素数でなくてはならない！
/*************************************/
Mod inv(const Mod a) { return a ^ (mo - 2); }
Mod operator/(const Mod a, const Mod b) { assert(b.num != 0); return a * inv(b); }
Mod operator/(const long long int a, const Mod b) { assert(b.num != 0); return Mod(a) * inv(b); }
Mod operator/=(Mod &a, const Mod b) { assert(b.num != 0); return a = a * inv(b); }

// n!と1/n!のテーブルを作る。
// nCrを高速に計算するためのもの。
//
// assertでnを超えていないかをきちんとテストすること。
//
// O(n log mo)
vector<Mod> fact, rfact;
void constructFactorial(const long long n) {
    fact.resize(n);
    rfact.resize(n);
    fact[0] = rfact[0] = 1;
    for (int i = 1; i < n; i++) {
        fact[i] = fact[i-1] * i;
        rfact[i] = Mod(1) / fact[i];
    }
}

// O(1)
Mod nCr(const long long n, const long long r) {
//    assert(n < (long long)fact.size());
    if (n < 0 || r < 0) return 0;
    return fact[n] * rfact[r] * rfact[n-r];
}

// O(k log mo) 
Mod nCrWithoutConstruction(const long long n, const long long k) {
    if (n < 0) return 0;
    if (k < 0) return 0;
    Mod ret = 1;
    for (int i = 0; i < k; i++) {
        ret *= n - (Mod)i;
        ret /= Mod(i+1);
    }
    return ret;
}
// n*mの盤面を左下から右上に行く場合の数
// O(1)
Mod nBm(const long long n, const long long m) {
    if (n < 0 || m < 0) return 0;
    return nCr(n + m, n);
}

/*************************************/
// GF(p)の行列演算
/*************************************/
using number = Mod;
using arr = vector<number>;
using matrix = vector<vector<Mod>>;

ostream &operator<<(ostream &o, const arr &v) { rep(i, v.size()) cout << v[i] << " "; return o; }
ostream &operator<<(ostream &o, const matrix &v) { rep(i, v.size()) cout << v[i]; return o; }

matrix zero(int n) { return matrix(n, arr(n, 0)); } // O(n^2)
matrix identity(int n) { matrix A(n, arr(n, 0)); rep(i, n) A[i][i] = 1; return A; } // O(n^2)
// O(n^2)
arr mul(const matrix &A, const arr &x) { 
    arr y(A.size(), 0); 
    rep(i, A.size()) rep(j, A[0].size()) y[i] += A[i][j] * x[j]; 
    return y; 
} 
// O(n^3)
matrix mul(const matrix &A, const matrix &B) {
    matrix C(A.size(), arr(B[0].size(), 0));
    rep(i, C.size())
        rep(j, C[i].size())
        rep(k, A[i].size())
        C[i][j] += A[i][k] * B[k][j];
    return C;
}
// O(n^2)
matrix plu(const matrix &A, const matrix &B) {
    matrix C(A.size(), arr(B[0].size(), 0));
    rep(i, C.size())
        rep(j, C[i].size())
            C[i][j] += A[i][j] + B[i][j];
    return C;
}
// O(n^2)
arr plu(const arr &A, const arr &B) {
    arr C(A.size());
    rep(i, A.size())
        C[i] += A[i] + B[i];
    return C;
}
// 構築なし累乗
//
// O(n^3 log e)
matrix pow(const matrix &A, long long e) {
    return e == 0 ? identity(A.size())  :
        e % 2 == 0 ? pow(mul(A, A), e/2) : mul(A, pow(A, e-1));
}
// 構築付き累乗
//
// powA: A^2^i
// O(n^3 log e)
matrix pow(const vector<matrix>& powA, long long e) { // powA[0]がA
//    cout << powA[0] << "^" << e <<endl;
    if (e <= 0) return identity(powA[0].size());
    matrix ret = identity(powA[0].size());
    rep(i, powA.size()) if (e & (1ll << i)) {
        ret = mul(ret, powA[i]);
    }
    return ret;
}
arr powmul(const vector<matrix>& powA, long long e, arr& a) { // powA[0]がA
//    cout << powA[0] << "^" << e <<endl;
    if (e <= 0) return a;
    arr ret = a;
    rep(i, powA.size()) if (e & (1ll << i)) {
        ret = mul(powA[i], ret);
    }
    return ret;
}

// Aを最大e乗まで計算するためのpowAを構築する。
// powAは副作用で返す
//
// O(n^3 log e)
void construct_powA(const matrix &A, long long e, vector<matrix>& powA) {
    powA.clear();
    powA.pb(A);
    for (int i = 1; (1ll << i) < e; i++) {
        powA.pb(mul(powA[i-1], powA[i-1]));
    }
}

// O(n)
number inner_product(const arr &a, const arr &b) {
    number ans = 0;
    for (int i = 0; i < (int)a.size(); ++i)
        ans += a[i] * b[i];
    return ans;
}
// O(n)
number tr(const matrix &A) {
    number ans = 0;
    for (int i = 0; i < (int)A.size(); ++i)
        ans += A[i][i];
    return ans;
}
// O( n^3 )
// modは素数でなければならない！！
number det(matrix A) {
    int n = A.size();
    assert(n == (int)A[0].size());
    number ans = 1;
    for (int i = 0; i < n; i++) {
        int pivot = -1;
        for (int j = i; j < n; j++)
            if (A[j][i]) {
                pivot = j;
                break;
            }
        if (pivot == -1) return 0;
        if (i != pivot) {
            swap(A[i], A[pivot]);
            ans *= -1;
        }
        number tmpinv = inv(A[i][i]);
        for (int j = i + 1; j < n; j++) {
            number c = A[j][i] * tmpinv;
            for (int k = i; k < n; k++) {
                A[j][k] = (A[j][k] - c * A[i][k]);
            }
        }
        ans *= A[i][i];
    }
    return ans;
}

// O( n^3 ).
// int rank(matrix A) はまだ

// O( n^3 ).
// modが2の時だけ使える演算
#define FOR(x,to) for(x=0;x<(to);x++) // repに変えちゃダメ。xがint xになると動かない
int gf2_rank(matrix A) { /* input */
    if (!A.size() || (A.size() && A[0].size())) return 0;
    int n = A.size();
    assert(mo == 2); 
    
    int i,j,k;
    FOR(i,n) {
        int be=i,mi=n+1;
        for(j=i;j<n;j++) {
            FOR(k,n) if(A[j][k]) break;
            if(k<mi) be=j,mi=k;
        }
        if(mi>=n) break;
        FOR(j,n) swap(A[i][j],A[be][j]);

        FOR(j,n) if(i!=j&&A[j][mi]) {
            FOR(k,n) A[j][k] += A[i][k]; // ^=のつもり
        }
    }
    return i;
}

/*************************************/
// 謎演算
/*************************************/

// gcdは関数__gcdを使いましょう。long long対応している。

// a x + b y = gcd(a, b)
long long extgcd(long long a, long long b, long long &x, long long &y) {
    long long g = a; x = 1; y = 0;
    if (b != 0) g = extgcd(b, a % b, y, x), y -= (a / b) * x;
    return g;
}

// 線型連立合同式 a[i] x == b[i] (mod m[i]) (i = 0, ..., n-1) を解く．
bool linearCongruences(const vector<long long> &a,
        const vector<long long> &b,
        const vector<long long> &m,
        long long &x, long long &M) {
    int n = a.size();
    x = 0; M = 1;
    rep(i, n) {
        long long a_ = a[i] % M, b_ = b[i] - a[i] * x, m_ = m[i];
        long long y, t, g = extgcd(a_, m_, y, t);
        if (b_ % g) return false;
        b_ /= g; m_ /= g;
        x += M * (y * b_ % m_);
        M *= m_;
    }
    x = (x + M) % M;
    return true;
}

// オイラーのφ関数
// LookUp Version
const int N = 1000000;
long long eulerPhi(long long n) {
    static int lookup = 0, p[N], f[N];
    if (!lookup) {
        rep(i,N) p[i] = 1, f[i] = i;
        for (int i = 2; i < N; ++i) {
            if (p[i]) {
                f[i] -= f[i] / i;
                for (int j = i+i; j < N; j+=i)
                    p[j] = 0, f[j] -= f[j] / i;
            }
        }
        lookup = 1;
    }
    return f[n];
}
template<int MOD>
struct ModInt {
	static const int Mod = MOD;
	unsigned x;
	ModInt() : x(0) {}
	ModInt(signed sig) { int sigt = sig % MOD; if(sigt < 0) sigt += MOD; x = sigt; }
	ModInt(signed long long sig) { int sigt = sig % MOD; if(sigt < 0) sigt += MOD; x = sigt; }
	int get() const { return (int)x; }

	ModInt &operator+=(ModInt that) { if((x += that.x) >= MOD) x -= MOD; return *this; }
	ModInt &operator-=(ModInt that) { if((x += MOD - that.x) >= MOD) x -= MOD; return *this; }
	ModInt &operator*=(ModInt that) { x = (unsigned long long)x * that.x % MOD; return *this; }
	ModInt &operator/=(ModInt that) { return *this *= that.inverse(); }

	ModInt operator+(ModInt that) const { return ModInt(*this) += that; }
	ModInt operator-(ModInt that) const { return ModInt(*this) -= that; }
	ModInt operator*(ModInt that) const { return ModInt(*this) *= that; }
	ModInt operator/(ModInt that) const { return ModInt(*this) /= that; }

	ModInt inverse() const {
		signed a = x, b = MOD, u = 1, v = 0;
		while(b) {
			signed t = a / b;
			a -= t * b; std::swap(a, b);
			u -= t * v; std::swap(u, v);
		}
		if(u < 0) u += Mod;
		ModInt res; res.x = (unsigned)u;
		return res;
	}

	bool operator==(ModInt that) const { return x == that.x; }
	bool operator!=(ModInt that) const { return x != that.x; }
	ModInt operator-() const { ModInt t; t.x = x == 0 ? 0 : Mod - x; return t; }
};
template<int MOD> ModInt<MOD> operator^(ModInt<MOD> a, unsigned long long k) {
	ModInt<MOD> r = 1;
	while(k) {
		if(k & 1) r *= a;
		a *= a;
		k >>= 1;
	}
	return r;
}
typedef ModInt<1000000007> mint;
typedef vector<mint> vmint;

struct RandomModInt {
    default_random_engine re;
    uniform_int_distribution<int> dist;
#ifndef _DEBUG
    RandomModInt() : re(random_device{}()), dist(1, mint::Mod - 1) { }
#else
    RandomModInt() : re(), dist(1, mint::Mod - 1) { }
#endif
    mint operator()() {
        mint r;
        r.x = dist(re);
        return r;
    }
} randomModInt;

void randomModIntVector(vector<mint> &v) {
    int n = (int)v.size();
    for(int i = 0; i < n; ++ i)
        v[i] = randomModInt();
}

ostream &operator<<(ostream &o, const mint v) {  o << v.x; return o; }

// GF(mo)列sから、それを生成する最小線形漸化式Cを復元する
//
// 入力: 漸化式が生成したGF(mo)列s
// 出力: d項間漸化式の係数C (size = d+1)
// 漸化式
//      C_0 s_{n} + C_1 s_{n-1} + ... + C_{L} s{n-L} = 0
// がsを生成した時、Cを求める。
//
// O(n^2)
//
// 例:
// s = [1, 2, 4, 8] -> C = [1, 1000000005(-2)] (s[1] - 2 * s[0] = 0)
// s = [1, 1, 1, 1] -> C = [1, 1000000006(-1)] (s[1] - s[0] = 0)
int berlekampMassey(const vector<mint> &s, vector<mint> &C) {
    int N = (int)s.size();
    C.assign(N + 1, mint());
    vector<mint> B(N + 1, mint());
    C[0] = B[0] = 1;
    int degB = 0;
    vector<mint> T;
    int L = 0, m = 1;
    mint b = 1;
    for(int n = 0; n < N; ++ n) {
        mint d = s[n];
        for(int i = 1; i <= L; ++ i)
            d += C[i] * s[n - i];
        if(d == mint()) {
            ++ m;
        } else {
            if(2 * L <= n)
                T.assign(C.begin(), C.begin() + (L + 1));
            mint coeff = -d * b.inverse();
            for(int i = -1; i <= degB; ++ i)
                C[m + i] += coeff * B[i];
            if(2 * L <= n) {
                L = n + 1 - L;
                B.swap(T);
                degB = (int)B.size() - 1;
                b = d;
                m = 1;
            } else {
                ++ m;
            }
        }
    }
    C.resize(L + 1);
    return L;
}

// GF(mo)列aから、それを生成する最小線形漸化式\phiを復元する
// berlekampMasseyとの違いは、係数の順序が違うのと安全用のassertチェックがあること。
//
// 入力: 漸化式が生成したGF(mo)列a
// 出力: d項間漸化式の係数\phi (size = d+1)
// 漸化式
//      \phi_0 a_{i} + \phi_1 a_{1} + ... + \phi_L a_L = 0
// がaを生成した時、\phiを求める。
//
// O(n^2)
//
// 例:
// s = [1, 2, 4, 8] -> C = [1000000005(-2), 1] (s[1] - 2 * s[0] = 0)
// s = [1, 1, 1, 1] -> C = [1000000006(-1), 1] (s[1] - s[0] = 0)
void computeMinimumPolynomialForLinearlyRecurrentSequence(const vector<mint> &a, vector<mint> &phi) {
    assert(a.size() % 2 == 0);
    int L = berlekampMassey(a, phi);
    reverse(phi.begin(), phi.begin() + (L + 1));
}

// 漸化式
//      \phi_0 a_{i} + \phi_1 a_{1} + ... + \phi_L a_L = 0
// と、initValues = a[0:phi.size()-1]が与えられる。
// この時、a[k]をinitValues(=a[0:phi.size()-1])の線形結合の係数を返す。
//     a[k] = coeff[0] * initValues[0] + coeff[1] * initValues[1] + ...  + coeff[d-1] * initValues[d-1] 
//
// O(n^2 log k)
void linearlyRecurrentSequenceCoeffs(long long k, const vector<mint> &phi_in, vector<mint> &coeffs) {
	int d = (int)phi_in.size() - 1;
	assert(d >= 0);
	assert(phi_in[d].get() == 1);

    coeffs = vector<mint>(d);
	vector<mint> square;
	coeffs[0] = 1;
	int l = 0;
	while ((k >> l) > 1) ++l;
	for (; l >= 0; --l) {
		square.assign(d * 2 - 1, mint());
        rep(i, d) rep(j, d) square[i + j] += coeffs[i] * coeffs[j];
		for (int i = d * 2 - 2; i >= d; -- i) {
			mint c = square[i];
			if (c.x == 0) continue;
            rep(j, d) square[i - d + j] -= c * phi_in[j];
		}
        rep(i, d)
			coeffs[i] = square[i];
		if (k >> l & 1) {
			mint lc = coeffs[d - 1];
			for(int i = d - 1; i >= 1; -- i)
				coeffs[i] = coeffs[i - 1] - lc * phi_in[i];
			coeffs[0] = mint() - lc * phi_in[0];
		}
	}
}

// 漸化式
//      \phi_0 a_{i} + \phi_1 a_{1} + ... + \phi_L a_L = 0
// と、initValues = a[0:phi.size()-1]が与えられる。
// この時、
//      a_{k}を求める
//
// O(n^2 log k)
// 
// また、副産物として、a[k]をinitVectorの線形結合として表す係数coeffが得られる
// a[k] = coeff[0] * initValues[0] + coeff[1] * initValues[1] + ...  + coeff[d-1] * initValues[d-1] 
// 
mint linearlyRecurrentSequenceValue(long long k, const vector<mint> &initValues, const vector<mint> &phi) {
    int d = phi.size() - 1;
	if(d == 0) return mint();
	assert(d <= (int)initValues.size());
    assert(k >= 0);

    if(k < (int)initValues.size())
        return initValues[(int)k];

    vector<mint> coeffs;
    linearlyRecurrentSequenceCoeffs(k, phi, coeffs);

	mint res; rep(i, d) res += coeffs[i] * initValues[i];
	return res;
}

// 線形漸化的数列aのk番目は？
// O(n^2 log k)
mint reconstruct(long long k,  vector<mint> a) {
    if (a.size() % 2) a.pop_back();
    vector<mint> a_first_half;
    rep(i, a.size() / 2)
        a_first_half.pb(a[i]);
    vector<mint> phi;
    computeMinimumPolynomialForLinearlyRecurrentSequence(a, phi);
    return linearlyRecurrentSequenceValue(k, a_first_half, phi);
}

matrix A = {
    {1,1,1,1,1,1,1,0},
    {1,1,1,1,1,0,0,0},
    {1,1,1,0,0,0,0,1},
    {1,1,0,1,1,0,0,0},
    {1,1,0,1,1,1,0,0},
    {1,0,0,0,1,1,0,1},
    {0,0,1,0,0,1,1,0},
    {1,0,0,0,0,0,0,1},
};
arr x_org = 
{1,1,1,1,1,1,1,0};

ll solve(ll n) {
    arr x = x_org;
    vll rets;
    rets.pb(6);
    rep(_, n-1) {
        x = mul(A, x);
        Mod ret = 0;
        rep(i, 8) if (i != 6) {
            ret += x[i];
        }
        rets.pb(ret);
    }
    Mod ret = 0;
    rep(i, rets.size()) {
        ret += rets[i] * (n - i);
    }
    return ret;
}

int main(void) {
    ll n; cin >> n;
    vector<mint> a;
    repi(i, 0, 100) {
        a.pb(solve(i));
    }
    cout << reconstruct(n, a) << endl; // めんどいのでBerlekamp-Masseyで復元
    return 0;
}