#include <iostream>
 
using namespace std;
 
#define MAX_V (15)
#define INF   (1<<29)
 
int V;
int cost[MAX_V][MAX_V];
int dp[1 << MAX_V][MAX_V];
 
int main() {
   
  cin >> V;
   
  for(int i = 0; i < V; i++)
    for(int j = 0; j < V; j++)
      cost[i][j] = INF;
   
  int m;
   
   
  cin >> m;
  for(int i = 0; i < m; i++) {
    int s, t, c;
    cin >> s >> t >> c;
    cost[s][t] = c;
  }
 
  int SMAX = 1<<V;
   
  for(int i = 0; i < SMAX; i++)
    for(int j = 0; j < V; j++)
      dp[i][j] = INF;
   
  dp[SMAX-1][0] = 0;
  for(int S = SMAX-2; S >= 0; S--)
    for(int i = 0; i < V; i++)
      for(int j = 0; j < V; j++)
    if(!(S >> j & 1))
      dp[S][i] = min(dp[S][i], dp[S|(1 << j)][j] + cost[i][j]);
     
  cout << ( dp[0][0] == INF ? -1 : dp[0][0] ) << endl;
   
  return 0;
}
### 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>
 
using namespace std;
 
#define MAX_V (15)
#define INF   (1<<29)
 
int V;
int cost[MAX_V][MAX_V];
int dp[1 << MAX_V][MAX_V];
 
int main() {
   
  cin >> V;
   
  for(int i = 0; i < V; i++)
    for(int j = 0; j < V; j++)
      cost[i][j] = INF;
   
  int m;
   
   
  cin >> m;
  for(int i = 0; i < m; i++) {
    int s, t, c;
    cin >> s >> t >> c;
    cost[s][t] = c;
  }
 
  int SMAX = 1<<V;
   
  for(int i = 0; i < SMAX; i++)
    for(int j = 0; j < V; j++)
      dp[i][j] = INF;
   
  dp[SMAX-1][0] = 0;
  for(int S = SMAX-2; S >= 0; S--)
    for(int i = 0; i < V; i++)
      for(int j = 0; j < V; j++)
    if(!(S >> j & 1))
      dp[S][i] = min(dp[S][i], dp[S|(1 << j)][j] + cost[i][j]);
     
  cout << ( dp[0][0] == INF ? -1 : dp[0][0] ) << endl;
   
  return 0;
}
```