#include<bits/stdc++.h>
#define rep(i,n)for(int i=0;i<n;i++)
#define INF 0x3f3f3f3f
using namespace std;

int V;
int d[15][15];
int dp[1 << 15][15];

int rec(int S, int v) {
	if (dp[S][v] >= 0)return dp[S][v];
	if (S == (1 << V) - 1 && v == 0)return dp[S][v] = 0;
	int res = INF;
	rep(u, V) {
		if (!(S >> u & 1))res = min(res, rec(S | 1 << u, u) + d[v][u]);
	}
	return dp[S][v] = res;
}
int main() {
	int E; scanf("%d%d", &V, &E);
	memset(d, 0x3f, sizeof(d));
	memset(dp, -1, sizeof(dp));
	rep(i, E) {
		int s, t, c; scanf("%d%d%d", &s, &t, &c);
		d[s][t] = c;
	}
	int ans = rec(0, 0);
	printf("%d\n", ans == INF ? -1 : ans);
}
### Prompt

Your task is to create 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
#include<bits/stdc++.h>
#define rep(i,n)for(int i=0;i<n;i++)
#define INF 0x3f3f3f3f
using namespace std;

int V;
int d[15][15];
int dp[1 << 15][15];

int rec(int S, int v) {
	if (dp[S][v] >= 0)return dp[S][v];
	if (S == (1 << V) - 1 && v == 0)return dp[S][v] = 0;
	int res = INF;
	rep(u, V) {
		if (!(S >> u & 1))res = min(res, rec(S | 1 << u, u) + d[v][u]);
	}
	return dp[S][v] = res;
}
int main() {
	int E; scanf("%d%d", &V, &E);
	memset(d, 0x3f, sizeof(d));
	memset(dp, -1, sizeof(dp));
	rep(i, E) {
		int s, t, c; scanf("%d%d%d", &s, &t, &c);
		d[s][t] = c;
	}
	int ans = rec(0, 0);
	printf("%d\n", ans == INF ? -1 : ans);
}
```