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

const long long MOD = 1000000007;
int N;
long long dp[101][5][5][5];

int main() {
  cin >> N;
  dp[0][0][0][0] = 1;
  for (int i = 1; i <= N; ++i) {
    for (int j = 0; j < 5; ++j) {
      for (int k = 0; k < 5; ++k) {
        for (int l = 0; l < 5; ++l) {
          for (int m = 1; m < 5; ++m) {
            if (k == 1 && l == 3 && m == 2) continue;
            if (k == 1 && l == 2 && m == 3) continue;
            if (k == 3 && l == 1 && m == 2) continue;
            if (j == 1 && k == 3 && m == 2) continue;
            if (j == 1 && l == 3 && m == 2) continue;
            dp[i][k][l][m] += dp[i-1][j][k][l];
            dp[i][k][l][m] %= MOD;
          }
        }
      }
    }
  }
  long long res = 0;
  for (int i = 1; i < 5; ++i) for (int j = 1; j < 5; ++j) for (int k = 1; k < 5; ++k) res += dp[N][i][j][k], res %= MOD;
  cout << res << endl;
}
### Prompt

Your task is to create a CPP solution to the following problem:

You are given an integer N. Find the number of strings of length N that satisfy the following conditions, modulo 10^9+7:

* The string does not contain characters other than `A`, `C`, `G` and `T`.
* The string does not contain `AGC` as a substring.
* The condition above cannot be violated by swapping two adjacent characters once.

Constraints

* 3 \leq N \leq 100

Input

Input is given from Standard Input in the following format:


N


Output

Print the number of strings of length N that satisfy the following conditions, modulo 10^9+7.

Examples

Input

3


Output

61


Input

4


Output

230


Input

100


Output

388130742

### Response

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

const long long MOD = 1000000007;
int N;
long long dp[101][5][5][5];

int main() {
  cin >> N;
  dp[0][0][0][0] = 1;
  for (int i = 1; i <= N; ++i) {
    for (int j = 0; j < 5; ++j) {
      for (int k = 0; k < 5; ++k) {
        for (int l = 0; l < 5; ++l) {
          for (int m = 1; m < 5; ++m) {
            if (k == 1 && l == 3 && m == 2) continue;
            if (k == 1 && l == 2 && m == 3) continue;
            if (k == 3 && l == 1 && m == 2) continue;
            if (j == 1 && k == 3 && m == 2) continue;
            if (j == 1 && l == 3 && m == 2) continue;
            dp[i][k][l][m] += dp[i-1][j][k][l];
            dp[i][k][l][m] %= MOD;
          }
        }
      }
    }
  }
  long long res = 0;
  for (int i = 1; i < 5; ++i) for (int j = 1; j < 5; ++j) for (int k = 1; k < 5; ++k) res += dp[N][i][j][k], res %= MOD;
  cout << res << endl;
}
```