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 g[16][16], n, i, j, k, v, e, f, t, c;
int dp[1 << 17][16];
int main() {
  scanf("%d%d", &v, &e);
  for (i = 0; i < v; i++)
    for (j = 0; j < v; j++) g[i][j] = 10000000;
  for (i = 1; i < (1 << 16); i++)
    for (j = 0; j < v; j++) dp[i][j] = 10000000;
  for (i = 0; i < e; g[f][t] = c, i++) scanf("%d%d%d", &f, &t, &c);
  for (i = 0; i < (1 << v); i++)
    for (j = 0; j < v; j++)
      for (k = 0; k < v; k++)
        if (g[j][k] < 10000000 && !(i & (1 << k)))
          dp[i + (1 << k)][k] =
              (dp[i + (1 << k)][k] < dp[i][j] + g[j][k] ? dp[i + (1 << k)][k]
                                                        : dp[i][j] + g[j][k]);
  printf("%d\n", dp[(1 << v) - 1][0] == 10000000 ? -1 : dp[(1 << v) - 1][0]);
  return 0;
}
