#include<bits/stdc++.h>
using namespace std;
#define ll long long
const ll mod =1e9+7;
int dp[110][5][5][5];
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==1&&c==0&&d==2)continue;
       if(b==0&&c==2&&d==1)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;
       }
       printf("%lld\n",ans);
   return 0;
}

### Prompt

In CPP, your task is to solve 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 ll long long
const ll mod =1e9+7;
int dp[110][5][5][5];
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==1&&c==0&&d==2)continue;
       if(b==0&&c==2&&d==1)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;
       }
       printf("%lld\n",ans);
   return 0;
}

```