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;
typedef long long ll;
int dp[1<<16][16];//dp[i][j]=すでに訪れた頂点の集合をi現在jにいる状態から残りのすべての頂点をめぐって頂点０に帰るようなパスの重みの総和の最小値
int d[16][16];
const int INF=100000000;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);

	int n,m;cin>>n>>m;
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++)d[i][j]=INF;
	}
	for(int i=0;i<m;i++){
		int a,b,c;cin>>a>>b>>c;
		d[a][b]=c;
	}


	for(int i=0;i<(1<<n);i++){
		fill(dp[i],dp[i]+n,INF);
	}
	dp[(1<<n)-1][0]=0;

	for(int S=(1<<n)-2;S>=0;S--){
		for(int v=0;v<n;v++){
			for(int u=0;u<n;u++){
				if(!(S>>u&1)){
					dp[S][v]=min(dp[S][v],dp[S|1<<u][u]+d[v][u]);
				}
			}
		}
	}
	if(dp[0][0]>=INF)cout<<-1<<endl;
	else cout<<dp[0][0]<<endl;
	
}

