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 MAX_N = 16;
const long long MAX_V = 1 << MAX_N;
const long long INF = (long long)1e18;
vector<vector<pair<long long, long long> > > g(MAX_N);
long long dp[MAX_N][MAX_V];
long long n, m;
long long solve(long long v, unsigned int S) {
  if (!(S & (1 << v))) return INF;
  if (dp[v][S]) return dp[v][S];
  if (v == 0 && S == 1) return 0;
  long long res = INF;
  for (pair<long long, long long> tp : g[v]) {
    long long u = tp.first, w = tp.second;
    res = min(res, w + solve(u, S ^ (1 << v)));
  }
  return dp[v][S] = res;
}
int main() {
  cin >> n >> m;
  if (n > 25) {
    cout << "This gonna take too long, lol\n";
    return 0;
  } else if (n == 0) {
    cout << "Don't be silly\n";
    return 0;
  }
  long long u, v, w, i;
  for (i = 0; i < m; i++) {
    cin >> u >> v >> w;
    g[u].push_back(pair<long long, long long>(v, w));
  }
  long long res = INF;
  unsigned int S = (1 << n) - 1;
  for (pair<long long, long long> tp : g[0]) {
    long long v = tp.first, w = tp.second;
    res = min(res, w + solve(v, S));
  }
  if (res == INF)
    cout << "-1\n";
  else
    cout << res << '\n';
}
