#include <iostream>
#include <cstdio>

using namespace std;

const long long M = 1e9 + 7;
long long n;
long long d[105][9][9][9];

long long f(long long p, long long q, long long r, long long s)
{
    long long re = 0;
    if(p == n)
    {
        return 1;
    }
    if(d[p][q][r][s]) return d[p][q][r][s];
    for(long long i = 0; i < 4; i++)
    {
        if(r == 0 && s == 1 && i == 2) continue;
        if(r == 0 && s == 2 && i == 1) continue;
        if(r == 1 && s == 0 && i == 2) continue;
        if(q == 0 && s == 1 && i == 2) continue;
        if(q == 0 && r == 1 && i == 2) continue;
        re += f(p + 1, r, s, i);
    }
    return d[p][q][r][s] = re % M;
}

int main()
{
    scanf("%lld", &n);
    printf("%lld", f(0, 4, 4, 4));
    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 <iostream>
#include <cstdio>

using namespace std;

const long long M = 1e9 + 7;
long long n;
long long d[105][9][9][9];

long long f(long long p, long long q, long long r, long long s)
{
    long long re = 0;
    if(p == n)
    {
        return 1;
    }
    if(d[p][q][r][s]) return d[p][q][r][s];
    for(long long i = 0; i < 4; i++)
    {
        if(r == 0 && s == 1 && i == 2) continue;
        if(r == 0 && s == 2 && i == 1) continue;
        if(r == 1 && s == 0 && i == 2) continue;
        if(q == 0 && s == 1 && i == 2) continue;
        if(q == 0 && r == 1 && i == 2) continue;
        re += f(p + 1, r, s, i);
    }
    return d[p][q][r][s] = re % M;
}

int main()
{
    scanf("%lld", &n);
    printf("%lld", f(0, 4, 4, 4));
    return 0;
}

```