#include<iostream>
#include<algorithm>

using namespace std;

int v, n, dis[15][15];
int dp[(1 << 15)][15];
const int INF = 100000000;

int main() {
	cin >> v >> n;
	for (int i = 0; i < v; i++)for (int j = 0; j < v; j++)dis[i][j] = INF;
	for (int i = 0; i < (1 << v); i++)for (int j = 0; j < v; j++)dp[i][j] = INF;
	for (int i = 0; i < n; i++) {
		int a, b, c;
		cin >> a >> b >> c;
		dis[a][b] = c;
	}
	dp[0][0] = 0;
	for (int i = 0; i < (1 << v) - 1; i++) {
		for (int j = 0; j < v; j++) {
			if (i && !(i & (1 << j)))continue;
			for (int k = 0; k < v; k++) {
				if (j == k)continue;
				if (!(i & (1 << k))) {
					dp[i | (1 << k)][k] = min(dp[i | (1 << k)][k], dp[i][j] + dis[j][k]);
				}
			}
		}
	}
	if (dp[(1 << v) - 1][0] == INF)cout << -1 << endl;
	else cout << dp[(1 << v) - 1][0] << endl;
	return 0;
}
### 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<iostream>
#include<algorithm>

using namespace std;

int v, n, dis[15][15];
int dp[(1 << 15)][15];
const int INF = 100000000;

int main() {
	cin >> v >> n;
	for (int i = 0; i < v; i++)for (int j = 0; j < v; j++)dis[i][j] = INF;
	for (int i = 0; i < (1 << v); i++)for (int j = 0; j < v; j++)dp[i][j] = INF;
	for (int i = 0; i < n; i++) {
		int a, b, c;
		cin >> a >> b >> c;
		dis[a][b] = c;
	}
	dp[0][0] = 0;
	for (int i = 0; i < (1 << v) - 1; i++) {
		for (int j = 0; j < v; j++) {
			if (i && !(i & (1 << j)))continue;
			for (int k = 0; k < v; k++) {
				if (j == k)continue;
				if (!(i & (1 << k))) {
					dp[i | (1 << k)][k] = min(dp[i | (1 << k)][k], dp[i][j] + dis[j][k]);
				}
			}
		}
	}
	if (dp[(1 << v) - 1][0] == INF)cout << -1 << endl;
	else cout << dp[(1 << v) - 1][0] << endl;
	return 0;
}
```