#include <bits/stdc++.h>
#define ll long long
#define INF 999999999
#define MOD 1000000007
#define rep(i,n) for(int i=0;i<n;i++)

using namespace std;

typedef pair<int,int>P;

const int MAX_N = 15;
const int MAX_M = 225;

int n,m;
ll dp[1 << MAX_N][MAX_N];
ll d[MAX_N][MAX_N];
int s[MAX_M];
int t[MAX_M];
int D[MAX_M];

ll solve()
{
	rep(S,1<<n){
		fill(dp[S],dp[S]+n,INF);
	}
	dp[(1<<n)-1][0] = 0;
	for(int S=(1<<n)-2;S>=0;S--){
		rep(v,n){
			if(S==0 || (S>>v&1)){
				rep(u,n){
					if(!(S>>u&1)){
						dp[S][v] = min(dp[S][v],dp[S|1<<u][u] + d[v][u]);
					}
				}
			}
		}
	}
	return dp[0][0];
}

int main()
{
	ll ans=INF-1;
	cin >> n >> m;
	rep(i,m){
		cin >> s[i] >> t[i] >> D[i];
	}
	rep(i,n){
		fill(d[0],d[n],INF);
		rep(j,m){
			d[(s[j]+n-i)%n][(t[j]+n-i)%n] = D[j];
		} 
		ans = min(ans,solve());
	}
	if(ans > 16000){
		cout << "-1\n";
	}else{
		cout << ans << "\n";
	}
}
### Prompt

Create a solution in Cpp for the following problem:

For a given weighted directed graph G(V, E), find the distance of the shortest route that meets the following criteria:

* It is a closed cycle where it ends at the same point it starts.
* It visits each vertex exactly once.

Constraints

* 2 ≤ |V| ≤ 15
* 0 ≤ di ≤ 1,000
* There are no multiedge

Input


|V| |E|
s0 t0 d0
s1 t1 d1
:
s|E|-1 t|E|-1 d|E|-1


|V| is the number of vertices and |E| is the number of edges in the graph. The graph vertices are named with the numbers 0, 1,..., |V|-1 respectively.

si and ti represent source and target vertices of i-th edge (directed) and di represents the distance between si and ti (the i-th edge).

Output

Print the shortest distance in a line. If there is no solution, print -1.

Examples

Input

4 6
0 1 2
1 2 3
1 3 9
2 0 1
2 3 6
3 2 4


Output

16


Input

3 3
0 1 1
1 2 1
0 2 1


Output

-1

### Response

```cpp
#include <bits/stdc++.h>
#define ll long long
#define INF 999999999
#define MOD 1000000007
#define rep(i,n) for(int i=0;i<n;i++)

using namespace std;

typedef pair<int,int>P;

const int MAX_N = 15;
const int MAX_M = 225;

int n,m;
ll dp[1 << MAX_N][MAX_N];
ll d[MAX_N][MAX_N];
int s[MAX_M];
int t[MAX_M];
int D[MAX_M];

ll solve()
{
	rep(S,1<<n){
		fill(dp[S],dp[S]+n,INF);
	}
	dp[(1<<n)-1][0] = 0;
	for(int S=(1<<n)-2;S>=0;S--){
		rep(v,n){
			if(S==0 || (S>>v&1)){
				rep(u,n){
					if(!(S>>u&1)){
						dp[S][v] = min(dp[S][v],dp[S|1<<u][u] + d[v][u]);
					}
				}
			}
		}
	}
	return dp[0][0];
}

int main()
{
	ll ans=INF-1;
	cin >> n >> m;
	rep(i,m){
		cin >> s[i] >> t[i] >> D[i];
	}
	rep(i,n){
		fill(d[0],d[n],INF);
		rep(j,m){
			d[(s[j]+n-i)%n][(t[j]+n-i)%n] = D[j];
		} 
		ans = min(ans,solve());
	}
	if(ans > 16000){
		cout << "-1\n";
	}else{
		cout << ans << "\n";
	}
}
```