#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i = 0; i < (int)n; i++)
using ll = long long;
const ll INF = 1e17;

int main(){
  int v, e;
  cin >> v >> e;
  vector<vector<ll>> d(v,vector<ll>(v,INF));
  rep(i,e) {
    ll s, t, a;
    cin >> s >> t >> a;
    d[s][t] = a;
  }
  vector<vector<ll>> dp(1<<v,vector<ll>(v,INF));
  dp[0][0] = 0;
  rep(i,1<<v) {
    rep(j,v) {
      rep(k,v) {
        if(!(i&(1<<k))) dp[i|(1<<k)][k] = min(dp[i|(1<<k)][k],dp[i][j]+d[j][k]);
      } 
    }
  }
  cout << (dp[(1<<v)-1][0] == INF? -1 : dp[(1<<v)-1][0]) << endl;
  return 0;
}

### Prompt

In CPP, your task is to solve 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,n) for(int i = 0; i < (int)n; i++)
using ll = long long;
const ll INF = 1e17;

int main(){
  int v, e;
  cin >> v >> e;
  vector<vector<ll>> d(v,vector<ll>(v,INF));
  rep(i,e) {
    ll s, t, a;
    cin >> s >> t >> a;
    d[s][t] = a;
  }
  vector<vector<ll>> dp(1<<v,vector<ll>(v,INF));
  dp[0][0] = 0;
  rep(i,1<<v) {
    rep(j,v) {
      rep(k,v) {
        if(!(i&(1<<k))) dp[i|(1<<k)][k] = min(dp[i|(1<<k)][k],dp[i][j]+d[j][k]);
      } 
    }
  }
  cout << (dp[(1<<v)-1][0] == INF? -1 : dp[(1<<v)-1][0]) << endl;
  return 0;
}

```