#include <iostream>
#include <cmath>
#include <string.h>
using namespace std;

int V,E;
int d[15][15] = {0};
int dp[1 << 15][15] = {0};
int inf = 1e8;

int rec(int S, int v){
	if(dp[S][v]>=0) return dp[S][v];
	if(S == (1<<V) -1 && v==0) return (dp[S][v] = 0);
	int res = inf;
	for(int u=0;u<V;u++){
		if(!(S>>u & 1)){
			res = min(res, rec(S | 1<<u, u) + d[v][u]);
		}
	}
	return (dp[S][v] = res);
}

int main(){
	cin >> V >> E;
	for(int i=0;i<V;i++){
		for(int j=0;j<V;j++){
			d[i][j] = inf;
		}
	}
	for(int i=0;i<E;i++){
		int a,b,w;
		cin >> a >> b >> w;
		d[a][b] = w;
	}
	memset(dp,-1,sizeof(dp));
	int ans = rec(0,0);
	if(ans!=inf) cout << ans << endl;
	else cout << -1 << endl;
}

### Prompt

Please create a solution in CPP to 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 <iostream>
#include <cmath>
#include <string.h>
using namespace std;

int V,E;
int d[15][15] = {0};
int dp[1 << 15][15] = {0};
int inf = 1e8;

int rec(int S, int v){
	if(dp[S][v]>=0) return dp[S][v];
	if(S == (1<<V) -1 && v==0) return (dp[S][v] = 0);
	int res = inf;
	for(int u=0;u<V;u++){
		if(!(S>>u & 1)){
			res = min(res, rec(S | 1<<u, u) + d[v][u]);
		}
	}
	return (dp[S][v] = res);
}

int main(){
	cin >> V >> E;
	for(int i=0;i<V;i++){
		for(int j=0;j<V;j++){
			d[i][j] = inf;
		}
	}
	for(int i=0;i<E;i++){
		int a,b,w;
		cin >> a >> b >> w;
		d[a][b] = w;
	}
	memset(dp,-1,sizeof(dp));
	int ans = rec(0,0);
	if(ans!=inf) cout << ans << endl;
	else cout << -1 << endl;
}

```