#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef pair<ll,ll> mp;
#define inf 1e9
int main(){
    int v;
    cin>>v;
    vector<vector<int> > g(v,vector<int>(v,inf) );
    int e;
    cin>>e;
    for(int i=0;i<e;i++){
	int s,t,d;
	cin>>s>>t>>d;
	g[s][t] = d;
    }
    vector<vector<int> > dp( 1<<v , vector<int>(v,inf) );
    dp[ (1<<v) -1 ][0] = 0;
    for( int i = (1<<v) -1 ; i >=  0 ; i-- ){
	for( int j = 0 ; j < v ; j++ ){ // from
	    for( int k = 0 ; k < v ; k++ ){ // to
		if( ( (i>>k) & 1 ) == 0 ){
		    dp[i][j] = min(dp[i][j] , dp[ i | (1<<k) ] [k] + g[k][j] );
		}
	    }
	}
    }
    if( dp[0][0] == inf ){
	cout<<-1<<endl;
    }else{
	cout<<dp[0][0]<<endl;
    }
    return 0;
}


### Prompt

Your challenge is to write 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<bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef pair<ll,ll> mp;
#define inf 1e9
int main(){
    int v;
    cin>>v;
    vector<vector<int> > g(v,vector<int>(v,inf) );
    int e;
    cin>>e;
    for(int i=0;i<e;i++){
	int s,t,d;
	cin>>s>>t>>d;
	g[s][t] = d;
    }
    vector<vector<int> > dp( 1<<v , vector<int>(v,inf) );
    dp[ (1<<v) -1 ][0] = 0;
    for( int i = (1<<v) -1 ; i >=  0 ; i-- ){
	for( int j = 0 ; j < v ; j++ ){ // from
	    for( int k = 0 ; k < v ; k++ ){ // to
		if( ( (i>>k) & 1 ) == 0 ){
		    dp[i][j] = min(dp[i][j] , dp[ i | (1<<k) ] [k] + g[k][j] );
		}
	    }
	}
    }
    if( dp[0][0] == inf ){
	cout<<-1<<endl;
    }else{
	cout<<dp[0][0]<<endl;
    }
    return 0;
}


```