#include <iostream>
#include <vector>
#include <algorithm>
#include <string.h>
#define INF (1 << 30)
#define ll long long
#define MAX_N 16
using namespace std;

ll V, E;
ll G[MAX_N][MAX_N];
ll dp[1 << MAX_N][MAX_N];

ll 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;
	}

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

int main() {
	cin.tie(0); ios::sync_with_stdio(false);
	cin >> V >> E;
	for (int i = 0; i < MAX_N;i++) {
		for (int j = 0; j < MAX_N;j++) {
			G[i][j] = INF;
		}
	}
	for (int i = 0; i < E; i++) {
		int s, t, d; cin >> s >> t >> d;
		G[s][t] = d;
	}
	memset(dp, -1, sizeof(dp));
	ll Ans = rec(0, 0);
	if (Ans == INF)cout << -1 << endl;
	else cout << Ans << endl;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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 <vector>
#include <algorithm>
#include <string.h>
#define INF (1 << 30)
#define ll long long
#define MAX_N 16
using namespace std;

ll V, E;
ll G[MAX_N][MAX_N];
ll dp[1 << MAX_N][MAX_N];

ll 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;
	}

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

int main() {
	cin.tie(0); ios::sync_with_stdio(false);
	cin >> V >> E;
	for (int i = 0; i < MAX_N;i++) {
		for (int j = 0; j < MAX_N;j++) {
			G[i][j] = INF;
		}
	}
	for (int i = 0; i < E; i++) {
		int s, t, d; cin >> s >> t >> d;
		G[s][t] = d;
	}
	memset(dp, -1, sizeof(dp));
	ll Ans = rec(0, 0);
	if (Ans == INF)cout << -1 << endl;
	else cout << Ans << endl;
}
```