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 = 1e9, MOD = 1e9 + 7,
          around[] = {0, 1, 1, -1, -1, 0, -1, 1, 0, 0};
const int vx[] = {1, 0, -1, 0}, vy[] = {0, 1, 0, -1};
const long long LINF = 1e18;
const long double PI = abs(acos(-1));
int n, m, dp[(1 << 16)][15];
vector<pair<int, int> > g[15];
int solve(int s) {
  for (int i = (0); i < ((1 << 16)); i++)
    for (int j = (0); j < (15); j++) dp[i][j] = INF;
  dp[0][s] = 0;
  for (int i = (0); i < ((1 << n)); i++) {
    for (int j = (0); j < (n); j++) {
      if (dp[i][j] == INF) continue;
      for (auto e : g[j]) {
        if ((i >> e.first) & 1) continue;
        dp[i | (1 << e.first)][e.first] =
            min(dp[i | (1 << e.first)][e.first], dp[i][j] + e.second);
      }
    }
  }
  return dp[(1 << n) - 1][s];
}
int main() {
  cin >> n >> m;
  for (int i = (0); i < (m); i++) {
    int a, b, c;
    cin >> a >> b >> c;
    g[a].push_back({b, c});
  }
  int mi = INF;
  for (int i = (0); i < (n); i++) mi = min(mi, solve(i));
  cout << (mi == INF ? -1 : mi) << endl;
  return 0;
}
