#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int INF = 100000;

const int MAX_V = 15;
int d[MAX_V][MAX_V];
int V;

int dp[1 << MAX_V][MAX_V];

int solve(int S, int v)
{
  if(dp[S][v] >= 0) return dp[S][v];

  if(S == (1 << V) - 1 && v == 0) return dp[S][v] = 0;

  int ans = INF;
  for(int u = 0; u < V; u++)
    if(!(S >> u & 1))
      ans = min(ans, solve(S | 1 << u, u) + d[v][u]);

  return dp[S][v] = ans;
}

int main()
{
  int E;
  scanf("%d %d", &V, &E);

  fill(&d[0][0], &d[MAX_V][0], INF);
  for(int i = 0; i < E; i++){
    int s, t, c;
    scanf("%d %d %d", &s, &t, &c);
    d[s][t] = c;
  }

  fill(&dp[0][0], &dp[1 << MAX_V][0], -1);
  int ans = solve(0, 0);
  if(ans != INF)
    printf("%d\n", ans);
  else
    printf("-1\n");
}
### Prompt

Develop a solution in Cpp to the problem described below:

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 <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int INF = 100000;

const int MAX_V = 15;
int d[MAX_V][MAX_V];
int V;

int dp[1 << MAX_V][MAX_V];

int solve(int S, int v)
{
  if(dp[S][v] >= 0) return dp[S][v];

  if(S == (1 << V) - 1 && v == 0) return dp[S][v] = 0;

  int ans = INF;
  for(int u = 0; u < V; u++)
    if(!(S >> u & 1))
      ans = min(ans, solve(S | 1 << u, u) + d[v][u]);

  return dp[S][v] = ans;
}

int main()
{
  int E;
  scanf("%d %d", &V, &E);

  fill(&d[0][0], &d[MAX_V][0], INF);
  for(int i = 0; i < E; i++){
    int s, t, c;
    scanf("%d %d %d", &s, &t, &c);
    d[s][t] = c;
  }

  fill(&dp[0][0], &dp[1 << MAX_V][0], -1);
  int ans = solve(0, 0);
  if(ans != INF)
    printf("%d\n", ans);
  else
    printf("-1\n");
}
```