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
#include <iostream>
#include <algorithm>
using namespace std;

int path[15][15];
int dp[1 << 15][15];
const int INF = 999999;

int main()
{
	for (int i = 0; i < 15; i++)
	{
		for (int j = 0; j < 15; j++)
		{
			path[i][j] = INF;
		}
	}
	for (int i = 0; i < (1 << 15); i++)
	{
		for (int j = 0; j < 15; j++)
		{
			dp[i][j] = INF;
		}
	}

	int V, E;
	cin >> V >> E;

	for (int i = 0; i < E; i++)
	{
		int s, t, d;
		cin >> s >> t >> d;
		path[s][t] = d;
	}

	dp[0][0] = 0;
	for (int i = 0; i < (1 << V); i++)
	{
		for (int j = 0; j < V; j++)
		{
			if (dp[i][j] == INF) continue;
			for (int k = 0; k < V; k++)
			{
				if (i & (1 << k)) continue;
				if (path[j][k] == INF) continue;
				dp[i + (1 << k)][k] = min(dp[i + (1 << k)][k], dp[i][j] + path[j][k]);
			}
		}
	}
	if (dp[(1 << V) - 1][0] == INF) cout << -1 << endl;
	else cout << dp[(1 << V) - 1][0] << endl;

	return 0;
}