#include<bits/stdc++.h>
using namespace std;
const long long md=1e9+7;
long long n,dp[111][10][10][10];
int main()
{
	cin>>n;
	dp[0][4][4][4]=1;
	for(int i=0;i<n;i++)
	{
		for(int x=0;x<5;x++)
		{
			for(int y=0;y<5;y++)
			{
				for(int z=0;z<5;z++)
				{
					for(int j=0;j<4;j++)
					{
						if(y==0 && z==1 && j==2)
						{
							continue;
						}
						if(y==0 && z==2 && j==1)
						{
							continue;
						}
						if(y==1 && z==0 && j==2)
						{
							continue;
						}
						if(x==0 && y==1 && j==2)
						{
							continue;
						}
						if(x==0 && z==1 && j==2)
						{
							continue;
						}
						dp[i+1][y][z][j]+=dp[i][x][y][z];
						dp[i+1][y][z][j]%=md;
					}
				}
			}
		}
	}
	long long ans=0;
	for(int x=0;x<5;x++)
	{
		for(int y=0;y<5;y++)
		{
			for(int z=0;z<5;z++)
			{
				ans+=dp[n][x][y][z];
			}
		}
	}
	cout<<ans%md<<endl;
	return 0;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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 md=1e9+7;
long long n,dp[111][10][10][10];
int main()
{
	cin>>n;
	dp[0][4][4][4]=1;
	for(int i=0;i<n;i++)
	{
		for(int x=0;x<5;x++)
		{
			for(int y=0;y<5;y++)
			{
				for(int z=0;z<5;z++)
				{
					for(int j=0;j<4;j++)
					{
						if(y==0 && z==1 && j==2)
						{
							continue;
						}
						if(y==0 && z==2 && j==1)
						{
							continue;
						}
						if(y==1 && z==0 && j==2)
						{
							continue;
						}
						if(x==0 && y==1 && j==2)
						{
							continue;
						}
						if(x==0 && z==1 && j==2)
						{
							continue;
						}
						dp[i+1][y][z][j]+=dp[i][x][y][z];
						dp[i+1][y][z][j]%=md;
					}
				}
			}
		}
	}
	long long ans=0;
	for(int x=0;x<5;x++)
	{
		for(int y=0;y<5;y++)
		{
			for(int z=0;z<5;z++)
			{
				ans+=dp[n][x][y][z];
			}
		}
	}
	cout<<ans%md<<endl;
	return 0;
}
```