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>
int main(void) {
  int v, e, st[15][15], i, j, k, l, d[15][1 << 15], ans = -1;
  scanf("%d%d", &v, &e);
  memset(st, -1, sizeof(st));
  for (i = 0; i < e; i++) {
    int s, t, dt;
    scanf("%d%d%d", &s, &t, &dt);
    st[s][t] = dt;
  }
  for (i = 0; i < v; i++) {
    memset(d, -1, sizeof(d));
    d[i][1 << i] = 0;
    while (1) {
      int flag = 1;
      for (j = 0; j < v; j++) {
        for (k = 0; k < 1 << v; k++) {
          for (l = 0; l < v; l++) {
            if (d[j][k] != -1 && st[j][l] != -1 &&
                (d[l][k | (1 << l)] == -1 ||
                 d[l][k | (1 << l)] > d[j][k] + st[j][l])) {
              flag = 0;
              d[l][k | (1 << l)] = d[j][k] + st[j][l];
            }
          }
        }
      }
      if (flag) {
        break;
      }
    }
    if (d[i][(1 << v) - 1] != -1 && (ans == -1 || d[i][(1 << v) - 1] < ans)) {
      ans = d[i][(1 << v) - 1];
    }
  }
  printf("%d\n", ans);
  return 0;
}
