/*** author: yuji9511 ***/
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using lpair = pair<ll, ll>;
const ll MOD = 1e9+7;
const ll INF = 1e18;
#define rep(i,m,n) for(ll i=(m);i<(n);i++)
#define rrep(i,m,n) for(ll i=(m);i>=(n);i--)
#define printa(x,n) for(ll i=0;i<n;i++){cout<<(x[i])<<" \n"[i==n-1];};
void print() {}
template <class H,class... T>
void print(H&& h, T&&... t){cout<<h<<" \n"[sizeof...(t)==0];print(forward<T>(t)...);}
ll power(ll x, ll n){
    if(n == 0) return 1LL;
    ll res = power(x * x % MOD, n/2);
    if(n % 2 == 1) res = res * x % MOD;
    return res;
}
typedef vector<ll> vec;
typedef vector<vec> mat;
mat mult(mat &A, mat &B){
    mat C(A.size(), vec(B[0].size(),0));
    rep(i,0,A.size()){
        rep(k,0,B.size()){
            rep(j,0,B[0].size()){
                C[i][j] += A[i][k] * B[k][j];
                C[i][j] %= MOD;
            }
        }
    }
    return C;
}

mat pow_mat(mat A, ll n){
    if(n == 1) return A;
    if(n % 2 == 0){
        mat B = pow_mat(A, n/2);
        return mult(B,B);
    }else{
        mat B = pow_mat(A, n-1);
        return mult(A,B); 
    }
}


void solve(){
    ll N;
    cin >> N;
    ll p[6] = {};
    p[0] = 1;
    ll nn = power(6, MOD-2);
    rep(i,1,6){
        rep(j,0,i){
            p[i] += p[j];
            p[i] %= MOD;
        }
        p[i] *= nn;
        p[i] %= MOD;
    }
    // printa(p, 6);
    if(N <= 5){
        print(p[N]);
        return;
    }
    mat A(6, vec(6,0));
    rep(i,0,6){
        A[0][i] = power(6, MOD-2);
    }
    rep(i,0,5){
        A[i+1][i] = 1;
    }
    mat ans = pow_mat(A, N-5);
    ll res = 0;
    rep(i,0,6){
        res += ans[0][i] * p[5-i];
        res %= MOD;
    }
    print(res);


    

}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    solve();
}