#include <bits/stdc++.h>
using namespace std;

constexpr int INF = 1e9;
void chmin (int &a, int b) {
	if (a > b) a = b;
}

int main() {
	int n, m;
	cin >> n >> m;
	vector<vector<int>> g(n, vector<int>(n, INF));
	for (int i = 0; i < m; i++) {
		int a, b, c;
		cin >> a >> b >> c;
		g[a][b] = c;
	}

	vector<vector<int>> dp(1 << n, vector<int>(n, INF));
	dp[0][0] = 0;
	
	for (int s = 0; s < (1 << n); s++) {
		for (int v = 0; v < n; v++) {
			for (int u = 0; u < n; u++) {
				if (u == v or (s & (1 << u))) continue;
				chmin(dp[s | (1 << u)][u], dp[s][v] + g[v][u]);
			}
		}
	}
	
	int ans = dp[(1 << n) - 1][0];
	if (ans == INF) ans = -1;
	
	cout << ans << '\n';
	
	return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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>
using namespace std;

constexpr int INF = 1e9;
void chmin (int &a, int b) {
	if (a > b) a = b;
}

int main() {
	int n, m;
	cin >> n >> m;
	vector<vector<int>> g(n, vector<int>(n, INF));
	for (int i = 0; i < m; i++) {
		int a, b, c;
		cin >> a >> b >> c;
		g[a][b] = c;
	}

	vector<vector<int>> dp(1 << n, vector<int>(n, INF));
	dp[0][0] = 0;
	
	for (int s = 0; s < (1 << n); s++) {
		for (int v = 0; v < n; v++) {
			for (int u = 0; u < n; u++) {
				if (u == v or (s & (1 << u))) continue;
				chmin(dp[s | (1 << u)][u], dp[s][v] + g[v][u]);
			}
		}
	}
	
	int ans = dp[(1 << n) - 1][0];
	if (ans == INF) ans = -1;
	
	cout << ans << '\n';
	
	return 0;
}

```