#include <bits/stdc++.h>
using namespace std;
#define rep(i, a) for(int i=0; i<(a); i++)
typedef long long ll;
const int inf = 1001001001;

int main()
{
  int n, m; cin >> n >> m;
  vector<vector<int>> tr(n, vector<int>(n, inf));
  rep(i, m) {
    int s, t, d; cin >> s >> t >> d;
    tr[s][t] = d;
  }
  vector<vector<int>> dp(1<<n, vector<int>(n, inf));
  dp[(1<<n)-1][0] = 0;

  for (int S=(1<<n)-1; S>=0; S--) {
    for (int v=0; v<n; v++) {
      for (int u=0; u<n; u++) {
        if (!(S>>u&1)) {
          dp[S][v] = min(dp[S][v], dp[S|1<<u][u] + tr[v][u]);
        }
      }
    }
  }
  if (dp[0][0] == inf) cout << -1 << endl;
  else cout << dp[0][0] << endl;

  return 0;
}


### Prompt

Your challenge is to write a Cpp solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
#define rep(i, a) for(int i=0; i<(a); i++)
typedef long long ll;
const int inf = 1001001001;

int main()
{
  int n, m; cin >> n >> m;
  vector<vector<int>> tr(n, vector<int>(n, inf));
  rep(i, m) {
    int s, t, d; cin >> s >> t >> d;
    tr[s][t] = d;
  }
  vector<vector<int>> dp(1<<n, vector<int>(n, inf));
  dp[(1<<n)-1][0] = 0;

  for (int S=(1<<n)-1; S>=0; S--) {
    for (int v=0; v<n; v++) {
      for (int u=0; u<n; u++) {
        if (!(S>>u&1)) {
          dp[S][v] = min(dp[S][v], dp[S|1<<u][u] + tr[v][u]);
        }
      }
    }
  }
  if (dp[0][0] == inf) cout << -1 << endl;
  else cout << dp[0][0] << endl;

  return 0;
}


```