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

const long long mod=1000000007LL;

long long dp[105][5][5][5]={};

int main(int argc, char const *argv[])
{
	int n;
	cin>>n;
	dp[0][0][0][0]=1;
	for (int i = 0; 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)||(k==3&&l==1&&m==2)||(k==1&&l==2&&m==3)||(j==1&&k==3&&m==2)||(j==1&&l==3&&m==2))
						{
							continue;
						}
						dp[i+1][k][l][m]+=dp[i][j][k][l];
						dp[i+1][k][l][m]%=mod;
					}
				}
			}
		}
	}
	long long ans=0LL;
	for (int i = 0; i < 5; ++i)
	{
		for (int j = 0; j < 5; ++j)
		{
			for (int k = 0; k < 5; ++k)
			{
				ans+=dp[n][i][j][k];
				ans%=mod;
			}
		}
	}
	cout<<ans<<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 mod=1000000007LL;

long long dp[105][5][5][5]={};

int main(int argc, char const *argv[])
{
	int n;
	cin>>n;
	dp[0][0][0][0]=1;
	for (int i = 0; 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)||(k==3&&l==1&&m==2)||(k==1&&l==2&&m==3)||(j==1&&k==3&&m==2)||(j==1&&l==3&&m==2))
						{
							continue;
						}
						dp[i+1][k][l][m]+=dp[i][j][k][l];
						dp[i+1][k][l][m]%=mod;
					}
				}
			}
		}
	}
	long long ans=0LL;
	for (int i = 0; i < 5; ++i)
	{
		for (int j = 0; j < 5; ++j)
		{
			for (int k = 0; k < 5; ++k)
			{
				ans+=dp[n][i][j][k];
				ans%=mod;
			}
		}
	}
	cout<<ans<<endl;
	return 0;
}

```