A - 费马小定理

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求：3^0 + 3^1 +...+ 3^(N) mod 1000000007

Input

输入一个数N(0 <= N <= 10^9)

Output

输出：计算结果

Sample Input

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Sample Output

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40

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#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue>
#include <cstring>

using namespace std;
typedef long long ll;
const int maxn = 1e2 + 5;
const int INF = (1 << 30);
const double inf = 1e-6;
const int mod = 1e9 + 7;
const int MOD = 1e9 + 7;
int t, n, m, ans, cnt, cc;
int f[maxn], vis[maxn][maxn];
char mp[maxn][maxn];
struct node {
    int x, y, steps;
} no[maxn];
ll FastPow(ll n, ll k) {
    ll ans = 1;
    while(k) {
        if(k & 1) {
            ans *= n;
            ans %= mod;
        }
        k >>= 1;
        n *= n;
        n %= mod;
    }
    return ans;
}
int main() {
    scanf("%d", &n);
    ll a = (FastPow(3, n + 1) - 1 + mod) % mod;
    ll b = (FastPow(2, mod - 2)) % mod;
    printf("%lldn", (a * b) % mod);
    return 0;
}