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;
int dp[16][16];
int V, E;
int v[16][16] = {0};
list<int> li;
list<int>::iterator lp;
int Max = -1;
int same(int x) {
  for (auto i = li.begin(); i != li.end();) {
    if (*i == x) return 0;
    ++i;
  }
  return 1;
}
void addsum(void) {
  int sum = 0;
  auto i = li.begin();
  auto j = i;
  ++j;
  do {
    sum += v[*i][*j];
    ++i;
    ++j;
  } while (j != li.end());
  sum += v[*i][0];
  if (Max == -1 || sum < Max) {
    Max = sum;
  }
}
void tansaku(list<int>::iterator p) {
  if (li.size() < V) {
    for (int i = 1; i < V; ++i) {
      if (*p != i && v[*p][i] != -1) {
        if (same(i)) {
          li.push_back(i);
          tansaku(++p);
          --p;
          li.pop_back();
        }
      }
    }
  } else {
    if (v[*p][0] != -1) {
      addsum();
    }
  }
}
int main() {
  fill(v[0], v[16], -1);
  int s, t, d;
  scanf("%d %d", &V, &E);
  for (int i = 0; i < E; ++i) {
    scanf("%d %d %d", &s, &t, &d);
    v[s][t] = d;
  }
  li.push_back(0);
  tansaku(li.begin());
  printf("%d\n", Max);
}
