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 <bits/stdc++.h>
using namespace std;

const int INF = 1e9;

int main() {
	int V, E;
	cin >> V >> E;
	vector<vector<int>> G(V, vector<int>(V, INF));
	for (int i=0;i<E;i++) {
		int s, t, d;
		cin >> s >> t >> d;
		G[s][t] = d;
	}

	int ans = INF;
	for (int s=0;s<V;s++) {
		vector<vector<int>> dp((1<<V), vector<int>(V, INF));
		dp[(1<<s)][s] = 0;
		for (int i=0;i<(1<<V);i++) {
			for (int j=0;j<V;j++) {
				for (int k=0;k<V;k++) {
					if (dp[i][j] != INF && G[j][k] != INF && (i&(1<<k)) == 0) {
						dp[i+(1<<k)][k] = min(dp[i+(1<<k)][k], dp[i][j]+G[j][k]);
					}
				}
			}
		}
		for (int i=0;i<V;i++) {
			if (G[i][s] != INF) {
				ans = min(ans, dp[(1<<V)-1][i]+G[i][s]);
			}
		}
	}
	if (ans == INF) cout << -1 << endl;
	else cout << ans << endl;
}
