#include <bits/stdc++.h>
using namespace std;
static const int64_t MOD = 1e9 + 7;
static const int T = 0, A = 1, G = 2, C = 3;
int n;
int64_t dp[101][4][4][4];

void add(int64_t &a, int64_t b) {
	a = (a+b) % MOD;
}

bool ng(int a, int b, int c) {
	return (a == A && b == G && c == C);
}

void solve() {
	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]);
	}
	int64_t sum = 0;
	for(int a = 0; a < 4; a++) for(int b = 0; b < 4; b++) for(int c = 0; c < 4; c++) {
		add(sum, dp[n][a][b][c]);
	}
	cout << sum << endl;

}

int main() {
	dp[0][0][0][0] = 1;
	cin >> n;
	solve();
	return 0;
}
### Prompt

Develop a solution in cpp 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;
static const int64_t MOD = 1e9 + 7;
static const int T = 0, A = 1, G = 2, C = 3;
int n;
int64_t dp[101][4][4][4];

void add(int64_t &a, int64_t b) {
	a = (a+b) % MOD;
}

bool ng(int a, int b, int c) {
	return (a == A && b == G && c == C);
}

void solve() {
	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]);
	}
	int64_t sum = 0;
	for(int a = 0; a < 4; a++) for(int b = 0; b < 4; b++) for(int c = 0; c < 4; c++) {
		add(sum, dp[n][a][b][c]);
	}
	cout << sum << endl;

}

int main() {
	dp[0][0][0][0] = 1;
	cin >> n;
	solve();
	return 0;
}
```