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;
int dp[1 << 15][15];
vector<pair<int, int>> ls[15];
int main() {
  int v, e;
  cin >> v >> e;
  for (int i = 0; i < e; i++) {
    int s, t, d;
    cin >> s >> t >> d;
    ls[s].push_back(make_pair(t, d));
  }
  for (int i = 0; i < 15; i++) {
    for (int j = 0; j < 1 << 15; j++) dp[j][i] = inf;
  }
  dp[0][0] = 0;
  for (int i = 0; i < (int)ls[0].size(); i++) {
    int to = ls[0][i].first, cost = ls[0][i].second;
    dp[1 << to][to] = cost;
  }
  for (int i = 1; i < 1 << v; i++) {
    for (int j = 0; j < v; j++) {
      cout << dp[i][j] << " ";
      if (dp[i][j] == inf) continue;
      for (int k = 0; k < (int)ls[j].size(); k++) {
        int to = ls[j][k].first, cost = ls[j][k].second;
        if (i & (1 << to)) continue;
        dp[i | (1 << to)][to] = min(dp[i | (1 << to)][to], dp[i][j] + cost);
      }
    }
    cout << endl;
  }
  if (dp[(1 << v) - 1][0] == inf)
    cout << -1 << endl;
  else
    cout << dp[(1 << v) - 1][0] << endl;
  return 0;
}
