#include <bits/stdc++.h>

using namespace std;

int main() {

	const long long INF = 1ll << 60;
	int v, e;
	cin >> v >> e;
	vector<vector<long long>> G(v, vector<long long>(v, INF));
	int in1, in2, in3;
	for (int i = 0; i < e; i++) {
		cin >> in1 >> in2 >> in3;
		G[in1][in2] = in3;
	}

	vector < vector <long long >> dp(1 << v, vector<long long>(v, INF));

	dp[0][0] = 0;

	for (int s = 0; s < (1 << v); s++) {

		for (int i = 0; i < v; i++) {

			for (int j = 0; j < v; j++) {

				if ((s & (1 << i)) == 0) {

					if (i != j) {
						dp[s | (1 << i)][j] = min(dp[s | (1 << i)][j]
														, dp[s][i] + G[i][j]);
					}

				}

			}

		}
		
	}

	if (dp[(1 << v) - 1][0] != INF) {
		cout << dp[(1 << v) - 1][0] << endl;
	}
	else { cout << -1 << endl; }

	return 0;

}

### Prompt

Generate 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>

using namespace std;

int main() {

	const long long INF = 1ll << 60;
	int v, e;
	cin >> v >> e;
	vector<vector<long long>> G(v, vector<long long>(v, INF));
	int in1, in2, in3;
	for (int i = 0; i < e; i++) {
		cin >> in1 >> in2 >> in3;
		G[in1][in2] = in3;
	}

	vector < vector <long long >> dp(1 << v, vector<long long>(v, INF));

	dp[0][0] = 0;

	for (int s = 0; s < (1 << v); s++) {

		for (int i = 0; i < v; i++) {

			for (int j = 0; j < v; j++) {

				if ((s & (1 << i)) == 0) {

					if (i != j) {
						dp[s | (1 << i)][j] = min(dp[s | (1 << i)][j]
														, dp[s][i] + G[i][j]);
					}

				}

			}

		}
		
	}

	if (dp[(1 << v) - 1][0] != INF) {
		cout << dp[(1 << v) - 1][0] << endl;
	}
	else { cout << -1 << endl; }

	return 0;

}

```