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

ll dp [101][4][4][4];
const ll mod=1e9+7;

void add(ll &a,ll b){
	a=(a+b)%mod;
}

bool ng(int a,int b,int c){
	return (a==1&&b==2&&c==3);
}

int main(){
	ll n;
	cin>>n;
	dp[0][0][0][0]=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(ng(b,c,d)||ng(c,b,d)||ng(b,d,c))continue;
						if(ng(a,c,d)||ng(a,b,d))continue;
						add(dp[i+1][b][c][d],dp[i][a][b][c]);
					}
				}
			}
		}
	}
	ll ans=0;
    for(int a=0; a<4; a++) for(int b=0; b<4; b++) for(int c=0; c<4; c++)add(ans, dp[n][a][b][c]);
    cout << ans << endl;
}

### 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;
typedef long long ll;

ll dp [101][4][4][4];
const ll mod=1e9+7;

void add(ll &a,ll b){
	a=(a+b)%mod;
}

bool ng(int a,int b,int c){
	return (a==1&&b==2&&c==3);
}

int main(){
	ll n;
	cin>>n;
	dp[0][0][0][0]=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(ng(b,c,d)||ng(c,b,d)||ng(b,d,c))continue;
						if(ng(a,c,d)||ng(a,b,d))continue;
						add(dp[i+1][b][c][d],dp[i][a][b][c]);
					}
				}
			}
		}
	}
	ll ans=0;
    for(int a=0; a<4; a++) for(int b=0; b<4; b++) for(int c=0; c<4; c++)add(ans, dp[n][a][b][c]);
    cout << ans << endl;
}

```