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 <bitset>

using namespace std;

const int INF = 1 << 30;
const int V_MAX = 15;

int V, E;
int dp[1 << V_MAX][V_MAX];
int dist[V_MAX][V_MAX];

int main() {
    fill(dp[0], dp[(1<<V_MAX)-1]+V_MAX, INF);
    fill(dist[0], dist[V_MAX-1]+V_MAX, INF);

    cin >> V >> E;
    int from, to, cost;
    for(int i=0;i<E;++i) {
        cin >> from >> to >> cost;
        dist[from][to] = cost;
    }

    dp[0][0] = 0;
    for(int bit=0;bit<(1<<V);++bit) {
        for(int i=0;i<V;++i) {
            if(dp[bit][i] == INF || (bit && !(bit & (1<<i)))) continue;
            for(int j=0;j<V;++j) {
                if((bit & (1<<j)) || dist[i][j] == INF) continue;
                if(dp[bit | (1<<j)][j] > dp[bit][i] + dist[i][j]) {
                    dp[bit | (1<<j)][j] = dp[bit][i] + dist[i][j];
                }
            }
        }
    }
    cout << (dp[(1<<V)-1][0] == INF ? -1 : dp[(1<<V)-1][0]) << endl;
}
