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

int main() {
    int n;cin>>n;
    int mod=1e9+7;
    map<string,int> dp[n+1];
    dp[0]["..."]=1;
    string a="ATGC";
    for (int i=0;i<n;i++)
        for (auto p:dp[i])
            for (int j=0;j<4;j++) {
                string t=p.first+a[j];
                int f=t.find("AGC")==string::npos;
                for (int k=0;k<3;k++) {
                    swap(t[k],t[k+1]);
                    f&=t.find("AGC")==string::npos;
                    swap(t[k],t[k+1]);
                }
                if (f)
                    dp[i+1][t.substr(1)]=(dp[i+1][t.substr(1)]+p.second)%mod;
            }
    int ans=0;
    for (auto p:dp[n])
        ans=(ans+p.second)%mod;
    cout<<ans<<endl;
    return 0;
}
### Prompt

Please create a solution in CPP 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 <bits/stdc++.h>
using namespace std;
typedef long long ll;

int main() {
    int n;cin>>n;
    int mod=1e9+7;
    map<string,int> dp[n+1];
    dp[0]["..."]=1;
    string a="ATGC";
    for (int i=0;i<n;i++)
        for (auto p:dp[i])
            for (int j=0;j<4;j++) {
                string t=p.first+a[j];
                int f=t.find("AGC")==string::npos;
                for (int k=0;k<3;k++) {
                    swap(t[k],t[k+1]);
                    f&=t.find("AGC")==string::npos;
                    swap(t[k],t[k+1]);
                }
                if (f)
                    dp[i+1][t.substr(1)]=(dp[i+1][t.substr(1)]+p.second)%mod;
            }
    int ans=0;
    for (auto p:dp[n])
        ans=(ans+p.second)%mod;
    cout<<ans<<endl;
    return 0;
}
```