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>
using namespace std;
const int INF = (1 << 30) - 1;
const int mxv = 16;
int V, E;
int adj[mxv][mxv], dp[(1 << mxv)][mxv], a, b, c;
int tsp(int v, int bitmask) {
  if (bitmask == (1 << V) - 1) return adj[v][0];
  if (dp[bitmask][v] < INF) return dp[bitmask][v];
  for (int i = 0; i < V; ++i)
    if (!(bitmask & (1 << i)))
      dp[bitmask][v] =
          min(dp[bitmask][v], adj[v][i] + tsp(i, bitmask | (1 << i)));
  return dp[bitmask][v];
}
int main(void) {
  scanf("%d%d", &V, &E);
  for (int i = 0; i < V; ++i)
    for (int j = 0; j < V; ++j) adj[i][j] = INF;
  for (int i = 0; i < (1 << V); ++i)
    for (int j = 0; j < V; ++j) dp[i][j] = INF;
  for (int i = 0; i < E; ++i) scanf("%d%d%d", &a, &b, &c), adj[a][b] = c;
  int sol = tsp(0, 1);
  printf("%d\n", (sol == INF) ? -1 : sol);
  return 0;
}
