#include<bits/stdc++.h>
using namespace std;
#define N 110
#define ll long long
ll dp[N][4][4][4];
ll mod=1e9+7;
int main()
{
    int n;
    scanf("%d",&n);
    dp[0][3][3][3]=1;
    for(int i=0;i<n;i++)
        for(int a=0;a<4;a++)
        for(int b=0;b<4;b++)
        for(int c=0;c<4;c++)
        for(int d=0;d<4;d++)
    {
        if(b==0&&c==1&&d==2)continue;
        if(b==0&&c==2&&d==1)continue;
        if(b==1&&c==0&&d==2)continue;
        if(a==0&&c==1&&d==2)continue;
        if(a==0&&b==1&&d==2)continue;
        dp[i+1][b][c][d]+=dp[i][a][b][c];
        dp[i+1][b][c][d]%=mod;
    }
    ll ans=0;
        for(int a=0;a<4;a++)
        for(int b=0;b<4;b++)
        for(int c=0;c<4;c++)
            ans+=dp[n][a][b][c],ans%=mod;
        cout<<ans;
    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>
using namespace std;
#define N 110
#define ll long long
ll dp[N][4][4][4];
ll mod=1e9+7;
int main()
{
    int n;
    scanf("%d",&n);
    dp[0][3][3][3]=1;
    for(int i=0;i<n;i++)
        for(int a=0;a<4;a++)
        for(int b=0;b<4;b++)
        for(int c=0;c<4;c++)
        for(int d=0;d<4;d++)
    {
        if(b==0&&c==1&&d==2)continue;
        if(b==0&&c==2&&d==1)continue;
        if(b==1&&c==0&&d==2)continue;
        if(a==0&&c==1&&d==2)continue;
        if(a==0&&b==1&&d==2)continue;
        dp[i+1][b][c][d]+=dp[i][a][b][c];
        dp[i+1][b][c][d]%=mod;
    }
    ll ans=0;
        for(int a=0;a<4;a++)
        for(int b=0;b<4;b++)
        for(int c=0;c<4;c++)
            ans+=dp[n][a][b][c],ans%=mod;
        cout<<ans;
    return 0;
}

```