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;
class edge {
 public:
  int t, d, lim;
  edge(int it, int id, int ilim) {
    t = it;
    d = id;
    lim = ilim;
  }
  bool operator>(edge a) { return lim > a.lim; }
  bool operator<(edge a) { return lim < a.lim; }
  bool operator>=(edge a) { return lim >= a.lim; }
  bool operator<=(edge a) { return lim <= a.lim; }
  bool operator==(edge a) { return lim == a.lim; }
};
vector<edge> e[20];
long long dp[16][1 << 16];
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int s, t, d, ti;
    cin >> s >> t >> d >> ti;
    e[s - 1].push_back(edge(t - 1, d, ti));
    e[t - 1].push_back(edge(s - 1, d, ti));
  }
  for (int i = 0; i < n; i++) {
    for (int j = (1 << n) - 1; j >= 0; j--) {
      dp[i][j] = 2e14;
    }
  }
  dp[0][0] = 0;
  for (int i = 0; i < (1 << n) - 1; i++) {
    for (int j = 0; j < n; j++) {
      if (dp[j][i] == 2e14) continue;
      for (int k = 0; k < (int)e[j].size(); k++) {
        if (dp[j][i] + e[j][k].d <= e[j][k].lim) {
          int next = i | (1 << e[j][k].t);
          if (next == i) continue;
          if (dp[e[j][k].t][next] > dp[j][i] + e[j][k].d) {
            dp[e[j][k].t][next] = dp[j][i] + e[j][k].d;
          }
        }
      }
    }
  }
  if (dp[0][(1 << n) - 1] == 2e14) {
    cout << -1 << endl;
    return 0;
  }
  cout << dp[0][(1 << n) - 1] << endl;
  return 0;
}
