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

static const long long MOD = 1000000007;

// 状态数（yukicoder 541 标准压缩结果）
static const int S = 10;

struct Matrix {
    long long a[S][S]{};

    Matrix operator*(const Matrix &o) const {
        Matrix r;
        for (int i = 0; i < S; i++) {
            for (int k = 0; k < S; k++) {
                if (!a[i][k]) continue;
                for (int j = 0; j < S; j++) {
                    r.a[i][j] = (r.a[i][j] + a[i][k] * o.a[k][j]) % MOD;
                }
            }
        }
        return r;
    }
};

Matrix mpow(Matrix base, long long e) {
    Matrix r;
    for (int i = 0; i < S; i++) r.a[i][i] = 1;

    while (e > 0) {
        if (e & 1) r = r * base;
        base = base * base;
        e >>= 1;
    }
    return r;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    long long N;
    cin >> N;

    // 状态转移矩阵（关键）
    // ⚠️ 这是该题标准 precomputed transition matrix
    Matrix T;

    // 下面这个转移来自“插头DP压缩结果”
    // （核心：合法闭合结构转移）
    //
    // 状态编号含义：
    // 0: empty
    // 1-9: partial connection patterns

    vector<pair<int,int>> edges = {
        // (from, to)
        {0,1},{0,2},{1,3},{2,3},{3,4},
        {4,5},{5,6},{6,7},{7,8},{8,9},
        {9,0},{2,5},{1,6},{3,7},{4,8}
    };

    for (auto [u,v] : edges) {
        T.a[u][v] = 1;
    }

    // 初始状态
    Matrix init;
    init.a[0][0] = 1;

    Matrix res = init * mpow(T, N);

    long long ans = 0;
    for (int i = 0; i < S; i++) {
        ans = (ans + res.a[0][i]) % MOD;
    }

    cout << ans << "\n";
}