#include <bits/stdc++.h>

using namespace std;

const int INF = 1 << 30;

int V, E;
vector< pair< int, int > > g[15];
int dp[1 << 15][15];

int main()
{
  cin >> V >> E;
  for(int i = 0; i < E; i++) {
    int s, t, d;
    cin >> s >> t >> d;
    g[s].emplace_back(t, d);
  }

  fill_n(*dp, (1 << 15) * 15, INF);
  dp[0][0] = 0;
  auto chmin = [](int &a, int b) { a = min(a, b); };

  for(int i = 0; i < (1 << V); i++) {
    for(int j = 0; j < V; j++) {
      if(dp[i][j] == INF) continue;
      for(auto &p : g[j]) {
        if((i >> p.first) & 1) continue;
        chmin(dp[i | (1 << p.first)][p.first], dp[i][j] + p.second);
      }
    }
  }

  if(dp[(1 << V) - 1][0] == INF) cout << -1 << endl;
  else cout << dp[(1 << V) - 1][0] << endl;
}
### Prompt

Create a solution in cpp for 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;

const int INF = 1 << 30;

int V, E;
vector< pair< int, int > > g[15];
int dp[1 << 15][15];

int main()
{
  cin >> V >> E;
  for(int i = 0; i < E; i++) {
    int s, t, d;
    cin >> s >> t >> d;
    g[s].emplace_back(t, d);
  }

  fill_n(*dp, (1 << 15) * 15, INF);
  dp[0][0] = 0;
  auto chmin = [](int &a, int b) { a = min(a, b); };

  for(int i = 0; i < (1 << V); i++) {
    for(int j = 0; j < V; j++) {
      if(dp[i][j] == INF) continue;
      for(auto &p : g[j]) {
        if((i >> p.first) & 1) continue;
        chmin(dp[i | (1 << p.first)][p.first], dp[i][j] + p.second);
      }
    }
  }

  if(dp[(1 << V) - 1][0] == INF) cout << -1 << endl;
  else cout << dp[(1 << V) - 1][0] << endl;
}
```