#include<bits/stdc++.h>
using namespace std;

const int INF = 1e8;

int rec(int bit, int v, vector<vector<int>> &dp, vector<vector<int>> &g){
	int n = dp.at(0).size();
	if(dp.at(bit).at(v)!=-1) return dp.at(bit).at(v);
	if(bit==(1<<v)) return dp.at(bit).at(v)=0;
	
	int dist = INF;
	int pbit = bit & ~(1<<v);
	for(int i=0; i<n; i++){
		if(!(pbit&(1<<i))) continue;
		if(dist > rec(pbit, i, dp, g)+g.at(i).at(v)) {
			dist = rec(pbit, i, dp, g)+g.at(i).at(v);
		}
	}
	return dp.at(bit).at(v) = dist;
}
int main() {
	
	int n,m;
	cin >> n >> m;
	vector<vector<int>> g(n+1,vector<int>(n+1,INF));
	for(int i=0; i<m; i++){
		int s,t,d;
		cin >> s >> t >> d;
		g.at(s).at(t) = d; 
	}
	vector<vector<int>> dp(1<<(n+1),vector<int>(n+1,-1));
	int ans{INF},cnt;
	for(int i=0; i<n; i++){
		auto dp1 = dp;
		auto g1 = g;
		for(int j=0; j<n+1; j++){
			for(int k=0; k<n+1; k++){
				if(j!=n&&k==n) g1.at(j).at(k) = INF;
				if(j==n) g1.at(j).at(k) = g.at(i).at(k); 
			}
		}
		ans = min(ans, rec((1<<(n+1))-1, i, dp1, g1));
		
	}
	cout << (ans==INF?-1:ans) << endl;
}

### Prompt

Please create a solution in Cpp 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;

const int INF = 1e8;

int rec(int bit, int v, vector<vector<int>> &dp, vector<vector<int>> &g){
	int n = dp.at(0).size();
	if(dp.at(bit).at(v)!=-1) return dp.at(bit).at(v);
	if(bit==(1<<v)) return dp.at(bit).at(v)=0;
	
	int dist = INF;
	int pbit = bit & ~(1<<v);
	for(int i=0; i<n; i++){
		if(!(pbit&(1<<i))) continue;
		if(dist > rec(pbit, i, dp, g)+g.at(i).at(v)) {
			dist = rec(pbit, i, dp, g)+g.at(i).at(v);
		}
	}
	return dp.at(bit).at(v) = dist;
}
int main() {
	
	int n,m;
	cin >> n >> m;
	vector<vector<int>> g(n+1,vector<int>(n+1,INF));
	for(int i=0; i<m; i++){
		int s,t,d;
		cin >> s >> t >> d;
		g.at(s).at(t) = d; 
	}
	vector<vector<int>> dp(1<<(n+1),vector<int>(n+1,-1));
	int ans{INF},cnt;
	for(int i=0; i<n; i++){
		auto dp1 = dp;
		auto g1 = g;
		for(int j=0; j<n+1; j++){
			for(int k=0; k<n+1; k++){
				if(j!=n&&k==n) g1.at(j).at(k) = INF;
				if(j==n) g1.at(j).at(k) = g.at(i).at(k); 
			}
		}
		ans = min(ans, rec((1<<(n+1))-1, i, dp1, g1));
		
	}
	cout << (ans==INF?-1:ans) << endl;
}

```