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

ll dp[105][4][4][4];

int main() {
  int N;
  cin>>N;

  dp[0][3][3][3]=1;
  for (int i=0; i<N; i++) {
    for (int c1=0; c1<4; c1++) { 
      for (int c2=0; c2<4; c2++) {
	for (int c3=0; c3<4; c3++) {
	  for (int a=0; a<4; a++) {
	    if (c2==0&&c1==1&&a==2
		||c3==0&&c1==1&&a==2
		||c2==1&&c1==0&&a==2
		||c2==0&&c1==2&&a==1
		||c3==0&&c2==1&&a==2) continue;
	    dp[i+1][c2][c1][a]+=dp[i][c3][c2][c1];
	    dp[i+1][c2][c1][a]%=1000000007;
	  }
	}
      }
    }
  }
  ll ans=0;
  for (int c3=0; c3<4; c3++) {
    for (int c2=0; c2<4; c2++) {
      for (int c1=0; c1<4; c1++) {
	ans+=dp[N][c3][c2][c1];
        ans%=1000000007;
      }
    }
  }
  cout<<ans<<endl;
  return 0;
}
### 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>
typedef long long ll;
using namespace std;

ll dp[105][4][4][4];

int main() {
  int N;
  cin>>N;

  dp[0][3][3][3]=1;
  for (int i=0; i<N; i++) {
    for (int c1=0; c1<4; c1++) { 
      for (int c2=0; c2<4; c2++) {
	for (int c3=0; c3<4; c3++) {
	  for (int a=0; a<4; a++) {
	    if (c2==0&&c1==1&&a==2
		||c3==0&&c1==1&&a==2
		||c2==1&&c1==0&&a==2
		||c2==0&&c1==2&&a==1
		||c3==0&&c2==1&&a==2) continue;
	    dp[i+1][c2][c1][a]+=dp[i][c3][c2][c1];
	    dp[i+1][c2][c1][a]%=1000000007;
	  }
	}
      }
    }
  }
  ll ans=0;
  for (int c3=0; c3<4; c3++) {
    for (int c2=0; c2<4; c2++) {
      for (int c1=0; c1<4; c1++) {
	ans+=dp[N][c3][c2][c1];
        ans%=1000000007;
      }
    }
  }
  cout<<ans<<endl;
  return 0;
}
```