#define _CRT_SECURE_NO_WARNINGS
#define _USE_MATH_DEFINES
//#include "MyMath.h"
//#include "MyDisjointset.h"
#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <functional>
#include <stdio.h>
#include <math.h>
#include <string.h>
using namespace std;
typedef pair<int, int> P; typedef long long int ll;
const int INF = 10000000;
int n, e;
int d[15][15], dp[1 << 15][15];
int main() {
	cin >> n >> e;
	fill(d[0], d[15], INF);
	fill(dp[0], dp[1 << 15], INF);
	for (int i = 0; i < e; i++) {
		int s, t, dis; cin >> s >> t >> dis;
		d[s][t] = dis;
	}
	dp[(1 << n) - 1][0] = 0;
	for (int i = (1 << n) - 2; i >= 0; i--)
		for (int j = 0; j < n; j++)
			for (int k = 0; k < n; k++)
				if (!(i >> k & 1))
					dp[i][j] = min(dp[i][j], dp[i | 1 << k][k] + d[j][k]);
	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
#define _CRT_SECURE_NO_WARNINGS
#define _USE_MATH_DEFINES
//#include "MyMath.h"
//#include "MyDisjointset.h"
#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <functional>
#include <stdio.h>
#include <math.h>
#include <string.h>
using namespace std;
typedef pair<int, int> P; typedef long long int ll;
const int INF = 10000000;
int n, e;
int d[15][15], dp[1 << 15][15];
int main() {
	cin >> n >> e;
	fill(d[0], d[15], INF);
	fill(dp[0], dp[1 << 15], INF);
	for (int i = 0; i < e; i++) {
		int s, t, dis; cin >> s >> t >> dis;
		d[s][t] = dis;
	}
	dp[(1 << n) - 1][0] = 0;
	for (int i = (1 << n) - 2; i >= 0; i--)
		for (int j = 0; j < n; j++)
			for (int k = 0; k < n; k++)
				if (!(i >> k & 1))
					dp[i][j] = min(dp[i][j], dp[i | 1 << k][k] + d[j][k]);
	if (dp[0][0] >= INF) cout << -1 << endl;
	else cout << dp[0][0] << endl;
	return 0;
}
```