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_V = 16;
const int MAX_E = (1 << MAX_V);
const int INF = 1 << 28;
int V, E;
int G[MAX_V][MAX_V];
int dp[MAX_E][MAX_V];
int dfs(int v, int visit) {
  if (dp[visit][v] != INF) return dp[visit][v];
  if (visit == (1 << V) - 1) return G[v][0];
  for (int i = 0; i < (int)(V); ++i) {
    if (visit & (1 << i)) continue;
    dp[visit][v] = min(dfs(i, (visit | (1 << i))) + G[v][i], dp[visit][v]);
  }
  return dp[visit][v];
}
int main() {
  cin >> V >> E;
  fill_n((int *)dp, sizeof(dp) / sizeof(int), INF);
  fill_n((int *)G, sizeof(G) / sizeof(int), INF);
  for (int i = 0; i < (int)(E); ++i) {
    int s, t, d;
    cin >> s >> t >> d;
    G[s][t] = d;
  }
  int visit = 0;
  int ans = dfs(0, (visit | 1));
  if (ans == INF)
    cout << -1 << endl;
  else
    cout << ans << endl;
  return 0;
}
