#include <iostream>
#include <vector>
using namespace std;
#define INF 1500000
typedef long long ll;
typedef pair<ll,ll> P;
ll d[16][16];
ll dp[1<<15][16];
int main(void){
    int n,e;
    cin>>n>>e;
    for(int i=0;i<n;i++)for(int j=0;j<n;j++)d[i][j]=INF;
    for(int i=0;i<e;i++){
        int s,t,c;
        cin>>s>>t>>c;
        d[s][t]=c;
    }
    for(int j=0;j<n;j++)for(int i=0;i<1<<n;i++)dp[i][j]=INF;
    dp[(1<<n)-1][0]=0;
    for(int i=(1<<n)-1;i>=0;i--){
        for(int j=0;j<n;j++){
            
                for(int k=0;k<n;k++){
                    if((i>>k&1)==0){
                        dp[i][j]=min(dp[i|1<<k][k]+d[j][k],dp[i][j]);
                    }
                }
            
        }
    }
    if(dp[0][0]>=INF)cout<<-1<<endl;
    else cout<<dp[0][0]<<endl;
}
### Prompt

Your task is to create 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 <iostream>
#include <vector>
using namespace std;
#define INF 1500000
typedef long long ll;
typedef pair<ll,ll> P;
ll d[16][16];
ll dp[1<<15][16];
int main(void){
    int n,e;
    cin>>n>>e;
    for(int i=0;i<n;i++)for(int j=0;j<n;j++)d[i][j]=INF;
    for(int i=0;i<e;i++){
        int s,t,c;
        cin>>s>>t>>c;
        d[s][t]=c;
    }
    for(int j=0;j<n;j++)for(int i=0;i<1<<n;i++)dp[i][j]=INF;
    dp[(1<<n)-1][0]=0;
    for(int i=(1<<n)-1;i>=0;i--){
        for(int j=0;j<n;j++){
            
                for(int k=0;k<n;k++){
                    if((i>>k&1)==0){
                        dp[i][j]=min(dp[i|1<<k][k]+d[j][k],dp[i][j]);
                    }
                }
            
        }
    }
    if(dp[0][0]>=INF)cout<<-1<<endl;
    else cout<<dp[0][0]<<endl;
}
```