#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<(int)(n);i++)
using namespace std;
using ll = long long ;
using P = pair<int,int> ;
using pll = pair<long long,long long>;
constexpr int INF = 1e9;
constexpr long long LINF = 1e17;
constexpr int MOD = 1000000007;
constexpr double PI = 3.14159265358979323846;

using vec = vector<ll> ;
using mat =  vector<vec>;

mat mul(mat &A, mat &B,int mod) {
        mat C(A.size(),vec(B[0].size()));
        for(int i=0;i<A.size();i++){
                for(int k=0;k<B.size();k++){
                        for(int j=0;j<B[0].size();j++){
                                C[i][j] = (C[i][j] + A[i][k]*B[k][j]) % mod;
                        }
                }
        }
        return C;
}

mat matpow(mat A,ll n,int mod){
        mat B(A.size(), vec(A.size()));
        for(int i=0;i<A.size();i++) B[i][i] = 1;
        while(n > 0){
                if(n & 1) B = mul(B,A,mod);
                A = mul(A,A,mod);
                n >>= 1;
        }
        return B;
}

/*最大２の3０乗までの数なら計算できる.繰り返し二乗法*/
long long modpow(long long x, long long n) {
    long long ret = 1;
    while (n > 0) {
        if (n & 1) ret = (ret * x) % MOD;  
        x = (x * x) % MOD;
        n >>= 1;  
    }
    return ret;
}


/*繰り返し二乗法でxのmod逆元を取る*/
long long inverse(long long x){
    return modpow(x,MOD-2);
}


int main(){
    ll n;
    cin >> n;
    vector<vector<ll>> A(6,vector<ll>(6,0));
    for(int i=1;i<6;i++){
        A[i][i-1] = 1;
    }
    ll inv = inverse(6);
    rep(i,6) A[0][i] = inv;
    /*
    rep(i,6){
        rep(j,6) cout << A[i][j] <<" " ;
        cout << endl;
    }
    */

    vector<ll> ve(6,0);
    ve[5] = 1;
    for(int i=4;i>=0;i--){
        rep(j,6){
            if(i+j<6){
                ve[i] += (inv * ve[i+j])%MOD;
                ve[i] %= MOD;
            }
        }
    }

    /*
    rep(i,6) cout << ve[i] << " ";
    cout << endl;
    */

    if(n<=5){
        cout << ve[5-n] << endl;
        return 0;
    }

    mat B = matpow(A,n-5,MOD);

    /*
    rep(i,6){
        rep(j,6) cout << B[i][j] <<" " ;
        cout << endl;
    }
    */

    ll ans = 0;
    rep(i,6){
        ans += (ve[i] * B[0][i])%MOD;
        ans %= MOD;
    }
    cout << ans << endl;

    return 0;
}