#include <bits/stdc++.h>
#define INF 100000000
using namespace std;

int main()
{
	int V, E;
	cin >> V >> E;
	vector<vector<int>> dp(1 << V, vector<int>(V, INF)), edge(V,vector<int>(V, INF));
	for (int i = 0, s, t, d; i < E; i++) {
		cin >> s >> t >> d;
		edge[s][t] = d;
	}
	dp[0][0] = 0;
	for (int i = 1; i < (1 << V); i++) {
		for (int j = 0; j < V; j++) {
			if (i & (1 << j)) {
				for (int k = 0; k < V; k++) {
					dp[i][j] = min(dp[i][j], dp[i - (1 << j)][k] + edge[k][j]);
				}
			}
		}
	}
	if (dp[(1 << V) - 1][0] == INF)
		cout << -1 << endl;
	else
		cout << dp[(1 << V) - 1][0] << endl;
	return 0;
}
### Prompt

Your challenge is to write a Cpp solution 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 <bits/stdc++.h>
#define INF 100000000
using namespace std;

int main()
{
	int V, E;
	cin >> V >> E;
	vector<vector<int>> dp(1 << V, vector<int>(V, INF)), edge(V,vector<int>(V, INF));
	for (int i = 0, s, t, d; i < E; i++) {
		cin >> s >> t >> d;
		edge[s][t] = d;
	}
	dp[0][0] = 0;
	for (int i = 1; i < (1 << V); i++) {
		for (int j = 0; j < V; j++) {
			if (i & (1 << j)) {
				for (int k = 0; k < V; k++) {
					dp[i][j] = min(dp[i][j], dp[i - (1 << j)][k] + edge[k][j]);
				}
			}
		}
	}
	if (dp[(1 << V) - 1][0] == INF)
		cout << -1 << endl;
	else
		cout << dp[(1 << V) - 1][0] << endl;
	return 0;
}
```