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 long long hmod1 = 999999937;
const long long hmod2 = 1000000000 + 9;
const long long INF = 1 << 30;
const long long mod = 1000000000 + 7;
const long long dx4[4] = {1, 0, -1, 0};
const long long dy4[4] = {0, 1, 0, -1};
const long long dx8[8] = {1, 1, 1, 0, 0, -1, -1, -1};
const long long dy8[8] = {0, 1, -1, 1, -1, 0, 1, -1};
const double pi = 3.141592653589793;
long long n, m;
struct edge {
  long long to, cost;
};
long long g[20][20];
long long dp[1 << 15][15];
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> n >> m;
  for (long long i = 0; i < (n); i++)
    for (long long j = 0; j < (n); j++) g[i][j] = INT_MAX;
  for (long long i = 0; i < (m); i++) {
    long long s, t, d;
    cin >> s >> t >> d;
    g[s][t] = d;
  }
  for (long long i = 0; i < (1 << n); i++)
    for (long long j = 0; j < (n); j++) dp[i][j] = INT_MAX / 2;
  long long start = 0;
  dp[1 << start][start] = 0;
  for (long long i = 0; i < (1 << n); i++) {
    for (long long j = 0; j < (n); j++) {
      if (i >> j & 1) {
        for (long long k = 0; k < (n); k++) {
          dp[i][j] = min(dp[i][j], dp[i ^ (1 << j)][k] + g[k][j]);
        }
      }
    }
  }
  long long ans = INT_MAX;
  for (long long j = 0; j < (n); j++)
    ans = min(ans, dp[(1 << n) - 1][j] + g[j][start]);
  cout << (ans == INT_MAX ? -1 : ans) << endl;
}
