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 MAX_N = 15;
const int INF = (int)1e8;
int V;
int dp[MAX_N][1 << MAX_N];
vector<pair<int, int>> G[MAX_N];
int start;
int tsp(unsigned S, int v) {
  if (dp[v][S] != -1) return dp[v][S];
  if (S == 0 && v == start) return 0;
  int res = INF;
  for (int i = 0; i < (int)G[v].size(); i++) {
    int next = G[v][i].first;
    int d = G[v][i].second;
    if (S >> next & 1) {
      res = min(res, d + tsp(S & ~(1 << next), next));
    }
  }
  return dp[v][S] = res;
}
int main() {
  int E;
  cin >> V >> E;
  while (E--) {
    int s, t, d;
    cin >> s >> t >> d;
    G[s].push_back(make_pair(t, d));
  }
  int res = INF;
  for (int i = 0; i < V; i++) {
    start = i;
    memset(dp, -1, sizeof(dp));
    res = min(res, tsp((1 << V) - 1, i));
  }
  if (res == INF)
    cout << "-1" << endl;
  else
    cout << res << endl;
}
