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 V, E, A[15][15], MEMO[15][32768], VISIT[15][32768];
void input(void);
void init_A(void);
void init_MEMO(void);
int DP(int dest, int VIA);
int TSP(void);
int main(void) {
  input();
  printf("%d\n", TSP());
  return 0;
}
int TSP(void) {
  int i, VIA_init, min = 987654321;
  for (i = 0; i < V; i++) {
    init_MEMO();
    VIA_init = (1 << V) - 1 & ~(1 << i);
    int cnd = DP(i, VIA_init);
    if (cnd < min) min = cnd;
  }
  if (min >= 987654321)
    return -1;
  else
    return min;
}
int DP(int dest, int VIA) {
  if (VIA == 0) return A[0][dest];
  if (VISIT[dest][VIA] == 1) return MEMO[dest][VIA];
  VISIT[dest][VIA] = 1;
  int i, min = 987654321;
  for (i = 1; i < V; i++) {
    if (A[i][dest] != -1 && (VIA & (1 << i)) != 0) {
      int cnd = A[i][dest] + DP(i, VIA & ~(1 << i));
      if (cnd < min) min = cnd;
    }
  }
  MEMO[dest][VIA] = min;
  return min;
}
void input(void) {
  scanf("%d%d", &V, &E);
  init_A();
  int i, s, t, d;
  for (i = 0; i < E; i++) {
    scanf("%d%d%d", &s, &t, &d);
    A[s][t] = d;
  }
}
void init_A(void) {
  int i, j;
  for (i = 0; i < V; i++)
    for (j = 0; j < V; j++) A[i][j] = -1;
}
void init_MEMO(void) {
  int i, j;
  for (i = 0; i < V; i++) {
    for (j = 0; j < (1 << V); j++) {
      MEMO[i][j] = 0;
      VISIT[i][j] = 0;
    }
  }
}
