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
using System;
using System.Text;
using System.Collections.Generic;
using System.Linq;
class Solve{
    public Solve(){}
    StringBuilder sb;
    public static int Main(){
        new Solve().Run();
        return 0;
    }
    void Run(){
        sb = new StringBuilder();
        Calc();
        Console.Write(sb.ToString());
    }
    void Calc(){
        string[] str = Console.ReadLine().Split(' ');
        int V = int.Parse(str[0]);
        int E = int.Parse(str[1]);
        long[,] cost = new long[V,V];
        for(int i=0;i<V;i++){
            for(int j=0;j<V;j++){
                cost[i,j] = 1000000;
            }
        }
        for(int i=0;i<E;i++){
            str = Console.ReadLine().Split(' ');
            int s = int.Parse(str[0]);
            int t = int.Parse(str[1]);
            int d = int.Parse(str[2]);
            cost[s,t] = d;
        }
        long[,] DP = new long[1<<V,V];
        for(int i=1;i<1<<V;i++){
            for(int j=0;j<V;j++){
                if(((1 << j) & i) != 0){
                    DP[i,j] = 1000000;
                }
                else{
                    long min = 1000000;
                    for(int k=0;k<V;k++){
                        if(((1 << k) & i) != 0){
                            min = Math.Min(min,DP[i-(1 << k),k]+cost[k,j]);
                        }
                    }
                    DP[i,j] = min;
                }
            }
        }
        {
            long min = 1000000;
            for(int i=1;i<V;i++){
                min = Math.Min(min,DP[(1<<V)-1-(1<<i),i]+cost[i,0]);
            }
            if(min == 1000000){
                min = -1;
            }
            sb.Append(min+"\n");
        }
    }
}