#include<iostream>
#include<vector>
#include<string>
#include<algorithm>	
#include<map>
#include<set>
#include<utility>
#include<cmath>
#include<cstring>
#include<queue>
#include<cstdio>
#include<sstream>
#define loop(i,a,b) for(int i=a;i<b;i++) 
#define rep(i,a) loop(i,0,a)
#define pb push_back
#define mp make_pair
#define all(in) in.begin(),in.end()
const double PI=acos(-1);
const double EPS=1e-8;
const int inf=1e8;
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> pii;
int main(){
	int n,m;
	cin>>n>>m;
	vvi cost(n,vi(n,inf));
	while(m--){
		int a,b,c;
		cin>>a>>b>>c;
		cost[a][b]=c;
	}
	vvi dp(1<<n,vi(n,inf));
	dp[(1<<n)-1][0]=0;
	for(int i=(1<<n)-2;i>=0;i--)rep(j,n){
		rep(k,n)if((i&1<<k)==0){
			dp[i][j]=min(dp[i][j],dp[i|1<<k][k]+cost[j][k]);
		}
	}
	if(dp[0][0]!=inf)cout<<dp[0][0]<<endl;
	else cout<<-1<<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<iostream>
#include<vector>
#include<string>
#include<algorithm>	
#include<map>
#include<set>
#include<utility>
#include<cmath>
#include<cstring>
#include<queue>
#include<cstdio>
#include<sstream>
#define loop(i,a,b) for(int i=a;i<b;i++) 
#define rep(i,a) loop(i,0,a)
#define pb push_back
#define mp make_pair
#define all(in) in.begin(),in.end()
const double PI=acos(-1);
const double EPS=1e-8;
const int inf=1e8;
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> pii;
int main(){
	int n,m;
	cin>>n>>m;
	vvi cost(n,vi(n,inf));
	while(m--){
		int a,b,c;
		cin>>a>>b>>c;
		cost[a][b]=c;
	}
	vvi dp(1<<n,vi(n,inf));
	dp[(1<<n)-1][0]=0;
	for(int i=(1<<n)-2;i>=0;i--)rep(j,n){
		rep(k,n)if((i&1<<k)==0){
			dp[i][j]=min(dp[i][j],dp[i|1<<k][k]+cost[j][k]);
		}
	}
	if(dp[0][0]!=inf)cout<<dp[0][0]<<endl;
	else cout<<-1<<endl;
}
```