#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<queue>
#include<stack>
#include<numeric>
using namespace std;
#define INF (1 << 30) - 1
#define LINF 1LL << 60
const int MOD = 1000000007;
using ll = long long;
using P = pair<int, int>;

int dp[1 << 20][20];

int dist[20][20];

int main(){
  int n, m;
  cin >> n >> m;
  fill((int*)dist, (int*)dist + sizeof(dist) / sizeof(int), INF);
  for(int i = 0; i < m; ++i){
  	int s, t, d;
  	cin >> s >> t >> d;
  	dist[s][t] = d;
  }

  for(int i = 0; i < (1 << 20); ++i){
  	for(int j = 0; j < 20; ++j){
  		dp[i][j] = INF;
  	}
  }

  dp[0][0] = 0; 
  for(int s = 0; s < (1 << n); ++s){
  	for(int v = 0; v < n; ++v){ 
  		if(!(s & (1 << v)))continue;
  		int prev_s = s - (1 << v); 
  		for(int u = 0; u < n; ++u){ 
   			dp[s][v] = min(dp[s][v], dp[prev_s][u] + dist[u][v]);
  		}
  	}
  }

  if(dp[(1 << n) - 1][0] == INF)cout << -1 << endl;
  else cout << dp[(1 << n) - 1][0] << endl;
  return 0;
}


### Prompt

Construct a cpp code solution to the problem outlined:

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<algorithm>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<queue>
#include<stack>
#include<numeric>
using namespace std;
#define INF (1 << 30) - 1
#define LINF 1LL << 60
const int MOD = 1000000007;
using ll = long long;
using P = pair<int, int>;

int dp[1 << 20][20];

int dist[20][20];

int main(){
  int n, m;
  cin >> n >> m;
  fill((int*)dist, (int*)dist + sizeof(dist) / sizeof(int), INF);
  for(int i = 0; i < m; ++i){
  	int s, t, d;
  	cin >> s >> t >> d;
  	dist[s][t] = d;
  }

  for(int i = 0; i < (1 << 20); ++i){
  	for(int j = 0; j < 20; ++j){
  		dp[i][j] = INF;
  	}
  }

  dp[0][0] = 0; 
  for(int s = 0; s < (1 << n); ++s){
  	for(int v = 0; v < n; ++v){ 
  		if(!(s & (1 << v)))continue;
  		int prev_s = s - (1 << v); 
  		for(int u = 0; u < n; ++u){ 
   			dp[s][v] = min(dp[s][v], dp[prev_s][u] + dist[u][v]);
  		}
  	}
  }

  if(dp[(1 << n) - 1][0] == INF)cout << -1 << endl;
  else cout << dp[(1 << n) - 1][0] << endl;
  return 0;
}


```