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>
void print_table(unsigned short *tbl, int row, int col) {
  int i, j;
  for (i = 0; i < row; i++) {
    for (j = 0; j < col; j++) {
      printf("%u ", tbl[i * col + j]);
    }
    printf("\n");
  }
  printf("\n");
}
int main() {
  unsigned short *memo;
  int dist[15][15];
  int i, j, s, t, d, V, E, state, before_state, before_dist, min_dist;
  unsigned short *score;
  scanf("%d %d", &V, &E);
  for (i = 0; i < V; i++) {
    for (j = 0; j < V; j++) {
      dist[i][j] = 0;
    }
  }
  for (i = 0; i < E; i++) {
    scanf("%d %d %d", &s, &t, &d);
    dist[s][t] = d;
  }
  memo = (unsigned short *)calloc((1 << (V - 1)) * (V - 1),
                                  sizeof(unsigned short));
  state = (1 << (V - 1)) - 1;
  for (i = 1; i < V; i++) {
    memo[state * (V - 1) + i - 1] = dist[0][i];
  }
  while (--state >= 0) {
    for (i = 0; i < V - 1; i++) {
      if (!(state & (1 << i))) {
        before_state = state | (1 << i);
        before_dist = memo[before_state * (V - 1) + i];
        if (before_dist == 0) continue;
        for (j = 0; j < V - 1; j++) {
          if (dist[i + 1][j + 1] > 0) {
            d = before_dist + dist[i + 1][j + 1];
            score = &memo[state * (V - 1) + j];
            if (*score == 0) {
              *score = d;
            } else if (d < *score) {
              *score = d;
            }
          }
        }
      }
    }
  }
  min_dist = 100000;
  for (i = 0; i < V - 1; i++) {
    if (!dist[i + 1][0] || !memo[i]) continue;
    d = memo[i] + dist[i + 1][0];
    if (d < min_dist) {
      min_dist = d;
    }
  }
  printf("%d\n", min_dist < 100000 ? min_dist : -1);
  free(memo);
  return 0;
}
