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;
vector<vector<int> > g, mas, reb;
int getbit(int mask, int num) { return (mask >> num) & 1; }
int addbit(int mask, int num) { return mask | (1 << num); }
int main() {
  int n, m, a, b, c;
  cin >> n >> m;
  int nn = (1 << n);
  reb.resize(n);
  for (int i = 0; i < n; i++) reb[i].resize(n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      reb[i][j] = -1;
    }
  }
  mas.resize(n);
  for (int i = 0; i < n; i++) mas[i].resize(nn);
  g.resize(n);
  for (int i = 0; i < m; i++) {
    cin >> a >> b >> c;
    g[a].push_back(b);
    reb[a][b] = c;
  }
  mas[0][1] = 1;
  int nextvert, nextmask;
  for (int mask = 1; mask < nn - 1; mask += 2) {
    for (int vert = 0; vert < n; vert++) {
      if (mas[vert][mask] != 0) {
        for (int e = 0; e < g[vert].size(); e++) {
          nextvert = g[vert][e];
          if (!getbit(mask, nextvert)) {
            nextmask = addbit(mask, nextvert);
            if (mas[nextvert][nextmask] == 0 ||
                mas[nextvert][nextmask] >
                    mas[vert][mask] + reb[vert][nextvert]) {
              mas[nextvert][nextmask] = mas[vert][mask] + reb[vert][nextvert];
            }
          }
        }
      }
    }
  }
  int mask = nn - 1;
  int mi = 1000000000;
  for (int vert = 0; vert < n; vert++) {
    if (reb[vert][0] != -1 && mas[vert][mask] != 0 &&
        mi > mas[vert][mask] + reb[vert][0]) {
      mi = mas[vert][mask] + reb[vert][0];
    }
  }
  if (mi != 1000000000) {
    cout << mi - 1;
  } else {
    cout << -1;
  }
  return 0;
}
