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

const int INF = 1e6;

int v, e;
int g[20][20];
int dp[(1 << 20)][20];

int rec(int s, int t) {
	if (dp[s][t] != 0) return dp[s][t];
	if (s == (1 << t)) return dp[s][t] = g[0][t];
	int ret = INF;
	int prev = s & ~(1 << t);
	for (int u = 0; u < v; u++) {
		if (!(prev & (1 << u))) continue;
		ret = min(ret, rec(prev, u) + g[u][t]);
	}
	return dp[s][t] = ret;
}

int main() {
	cin >> v >> e;
	rep(i, 20) rep(j, 20) g[i][j] = INF;
	rep(i, e) {
		int s, t, d;
		cin >> s >> t >> d;
		g[s][t] = d;
	}
	int ans = rec((1 << v) - 1, 0);
	if (ans != INF) cout << ans << endl;
	else cout << -1 << 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 <bits/stdc++.h>
#define rep(i, n) for(int i = 0; i < (n); i++)
using namespace std;
using ll = long long;

const int INF = 1e6;

int v, e;
int g[20][20];
int dp[(1 << 20)][20];

int rec(int s, int t) {
	if (dp[s][t] != 0) return dp[s][t];
	if (s == (1 << t)) return dp[s][t] = g[0][t];
	int ret = INF;
	int prev = s & ~(1 << t);
	for (int u = 0; u < v; u++) {
		if (!(prev & (1 << u))) continue;
		ret = min(ret, rec(prev, u) + g[u][t]);
	}
	return dp[s][t] = ret;
}

int main() {
	cin >> v >> e;
	rep(i, 20) rep(j, 20) g[i][j] = INF;
	rep(i, e) {
		int s, t, d;
		cin >> s >> t >> d;
		g[s][t] = d;
	}
	int ans = rec((1 << v) - 1, 0);
	if (ans != INF) cout << ans << endl;
	else cout << -1 << endl;
	return 0;
}

```