#include <iostream>
#include <cmath>
#define INF 1e+9
using namespace std;

int n;
int cost[15][15];
int dp[1 << 15][15];

int rec(int S,int v){
	if(dp[S][v] >= 0) return dp[S][v];
	if(S == (1 << n) - 1 && v == 0) return dp[S][v] = 0;
	int res = INF;
	for(int i = 0;i < n;i++){
		if(!(S >> i & 1)){
			res = min(res,rec(S | 1 << i,i) + cost[v][i]);
		}
	}
	return dp[S][v] = res;
}

int main(){
	int e;
	cin >> n >> e;
	for(int i = 0;i < n;i++){
		for(int j = 0;j < n;j++) cost[i][j] = INF;
	}
	for(int i = 0;i < e;i++){
		int s,t,d;
		cin >> s >> t >> d;
		cost[s][t] = d;
	}
	for(int i = 0;i < (1 << n);i++){
		for(int j = 0;j < n;j++) dp[i][j] = -1;
	}
	int res = rec(0,0);
	if(res == INF) cout << -1 << endl;
	else cout << res << endl;
	return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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>
#define INF 1e+9
using namespace std;

int n;
int cost[15][15];
int dp[1 << 15][15];

int rec(int S,int v){
	if(dp[S][v] >= 0) return dp[S][v];
	if(S == (1 << n) - 1 && v == 0) return dp[S][v] = 0;
	int res = INF;
	for(int i = 0;i < n;i++){
		if(!(S >> i & 1)){
			res = min(res,rec(S | 1 << i,i) + cost[v][i]);
		}
	}
	return dp[S][v] = res;
}

int main(){
	int e;
	cin >> n >> e;
	for(int i = 0;i < n;i++){
		for(int j = 0;j < n;j++) cost[i][j] = INF;
	}
	for(int i = 0;i < e;i++){
		int s,t,d;
		cin >> s >> t >> d;
		cost[s][t] = d;
	}
	for(int i = 0;i < (1 << n);i++){
		for(int j = 0;j < n;j++) dp[i][j] = -1;
	}
	int res = rec(0,0);
	if(res == INF) cout << -1 << endl;
	else cout << res << endl;
	return 0;
}
```