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;
inline int getbit(int mask, int v) { return (mask >> v) & 1; }
inline int newmask(int mask, int v) { return mask | (1 << v); }
int main() {
  int n, m;
  cin >> n >> m;
  int cost[n][n];
  vector<int> g[n];
  int x, y, c;
  for (int i = 0; i < m; i++) {
    cin >> x >> y >> c;
    g[x].push_back(y);
    cost[x][y] = c;
  }
  int P = pow(2, n);
  int d[P][n];
  for (int i = 1; i < P; i++) {
    for (int j = 0; j < n; j++) {
      d[i][j] = -1;
    }
  }
  d[1][0] = 0;
  const int INF = 1e9;
  int ans = INF;
  for (int mask = 1; mask < P; mask++) {
    for (int v = 0; v < n; v++) {
      if (d[mask][v] != -1) {
        for (int i = 0; i < (int)g[v].size(); i++) {
          int u = g[v][i];
          if (mask < P - 1) {
            if (!getbit(mask, u)) {
              d[newmask(mask, u)][u] = d[mask][v] + cost[v][u];
            }
          } else {
            if (u == 0) {
              ans = min(ans, d[mask][v] + cost[v][0]);
            }
          }
        }
      }
    }
  }
  if (ans == INF) {
    cout << -1 << endl;
  } else {
    cout << ans << endl;
  }
}
