#include using namespace std; #define rep(i,n) for(ll i=0;i=0;i--) #define perl(i,r,l) for(ll i=r-1;i>=l;i--) #define fi first #define se second #define pb push_back #define ins insert #define pqueue(x) priority_queue,greater> #define all(x) (x).begin(),(x).end() #define CST(x) cout<> #define rev(x) reverse(x); using ll=long long; using vl=vector; using vvl=vector>; using pl=pair; using vpl=vector; using vvpl=vector; const ll MOD=1000000007; const ll MOD9=998244353; const int inf=1e9+10; const ll INF=4e18; const ll dy[9]={0,1,0,-1,1,1,-1,-1,0}; const ll dx[9]={1,0,-1,0,1,-1,1,-1,0}; template inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; } template inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; } const int mod = MOD; const int max_n = 200005; struct mint { ll x; // typedef long long ll; mint(ll x=0):x((x%mod+mod)%mod){} mint operator-() const { return mint(-x);} mint& operator+=(const mint a) { if ((x += a.x) >= mod) x -= mod; return *this; } mint& operator-=(const mint a) { if ((x += mod-a.x) >= mod) x -= mod; return *this; } mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;} mint operator+(const mint a) const { return mint(*this) += a;} mint operator-(const mint a) const { return mint(*this) -= a;} mint operator*(const mint a) const { return mint(*this) *= a;} mint pow(ll t) const { if (!t) return 1; mint a = pow(t>>1); a *= a; if (t&1) a *= *this; return a; } bool operator==(const mint &p) const { return x == p.x; } bool operator!=(const mint &p) const { return x != p.x; } // for prime mod mint inv() const { return pow(mod-2);} mint& operator/=(const mint a) { return *this *= a.inv();} mint operator/(const mint a) const { return mint(*this) /= a;} }; istream& operator>>(istream& is, mint& a) { return is >> a.x;} ostream& operator<<(ostream& os, const mint& a) { return os << a.x;} using vm=vector; using vvm=vector; struct combination { vector fact, ifact; combination(int n):fact(n+1),ifact(n+1) { assert(n < mod); fact[0] = 1; for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i; ifact[n] = fact[n].inv(); for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i; } mint operator()(int n, int k) { if (k < 0 || k > n) return 0; return fact[n]*ifact[k]*ifact[n-k]; } }comb(max_n); vm conv(vm a,vm b){ vm c(a.size()+b.size()-1); rep(i,a.size()){ rep(j,b.size()){ c[i+j]+=a[i]*b[j]; } } return c; } vector BerlekampMassey(const vector &s) { const int N = (int)s.size(); vector b, c; b.reserve(N + 1); c.reserve(N + 1); b.push_back(mint(1)); c.push_back(mint(1)); mint y = mint(1); for (int ed = 1; ed <= N; ed++) { int l = int(c.size()), m = int(b.size()); mint x = 0; for (int i = 0; i < l; i++) x += c[i] * s[ed - l + i]; b.emplace_back(mint(0)); m++; if (x == mint(0)) continue; mint freq = x / y; if (l < m) { auto tmp = c; c.insert(begin(c), m - l, mint(0)); for (int i = 0; i < m; i++) c[m - 1 - i] -= freq * b[m - 1 - i]; b = tmp; y = x; } else { for (int i = 0; i < m; i++) c[l - 1 - i] -= freq * b[m - 1 - i]; } } reverse(begin(c), end(c)); return c; } template vector kitamasa(vector Q,vector a) { assert(!Q.empty() && Q[0] != 0); assert((int)a.size() >= int(Q.size()) - 1); vector P(Q.size()*2-2); for(ll i=0;i struct bostan_mori { vector p, q; bostan_mori(vector &_p, vector &_q) : p(_p), q(_q) {} void rever(vector &f) const { int d = f.size(); rep(i, d) if (i&1) f[i] = -f[i]; } void even(vector &f) const { int d = (f.size() + 1) >> 1; rep(i, d) f[i] = f[i<<1]; f.resize(d); } void odd(vector &f) const { int d = f.size() >> 1; rep(i, d) f[i] = f[i<<1|1]; f.resize(d); } vector convolution(vector a,vector b) const{ int n=a.size(),m=b.size(); vector c(n+m-1); rep(i,n)rep(j,m)c[i+j]+=a[i]*b[j]; return c; } T operator[] (ll n) const { vector _p(p), _q(q), _q_rev(q); rever(_q_rev); for (; n; n >>= 1) { _p = convolution(move(_p), _q_rev); if (n&1) odd(_p); else even(_p); _q = convolution(move(_q), move(_q_rev)); even(_q); _q_rev = _q; rever(_q_rev); } return _p[0] / _q[0]; } }; //https://nyaannyaan.github.io/library/fps/kitamasa.hpp //https://atcoder.jp/contests/tdpc/submissions/34362182 //線形漸化式のprefixからn項目を復元できる。 bostan_mori interpolation(vm a){ auto q=BerlekampMassey(a); auto p=kitamasa(q,a); return bostan_mori(p,q); } int main(){ ll n,p,c;cin >> n >> p >> c; vl prime={2,3,5,7,11,13}; vl compos={4,6,8,9,10,12}; vvm dp(301,vm(4000));dp[0][0]=1; { rep(i,6){ rep(j,300){ rep(k,4000){ if(k+prime[i]<4000)dp[j+1][k+prime[i]]+=dp[j][k]; } } } } vvm ndp(301,vm(4000));ndp[0][0]=1; { rep(i,6){ rep(j,300){ rep(k,4000){ if(k+compos[i]<4000)ndp[j+1][k+compos[i]]+=ndp[j][k]; } } } } auto f=dp[p];auto g=ndp[c]; f=conv(f,g); while(f.back().x==0)f.pop_back(); { ll m=f.size()*2; vm naive(m); rep(i,f.size())naive[i]=1; for(ll j=f.size();j