#include <vector>
#include <iostream>
#include <algorithm>

using namespace std;

int main()
{
	int V, E, s, t, d;

	cin >> V >> E;

	vector<vector<int> > M(V, vector<int>(V, 1 << 27));

	for (int i = 0; i < E; i++)
	{
		cin >> s >> t >> d;

		M[s][t] = d;
	}

	vector<vector<int> > dp(V, vector<int>(1 << V, 1 << 27));

	dp[0][0] = 0;

	for (int i = 0; i < V; i++)
	{
		for (int j = 0; j < V; j++)
		{
			for (int k = 0; k < (1 << V); k++)
			{
				for (int l = 0; l < V; l++)
				{
					if ((k / (1 << j)) % 2 == 1)
					{
						dp[j][k] = min(dp[j][k], dp[l][k - (1 << j)] + M[l][j]);
					}
				}
			}
		}
	}

	cout << (dp[0][(1 << V) - 1] != (1 << 27) ? dp[0][(1 << V) - 1] : -1) << endl;

	return 0;
}

// Time : O ( V ^ 3 * 2 ^ V ) = 110,592,000
### 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 <vector>
#include <iostream>
#include <algorithm>

using namespace std;

int main()
{
	int V, E, s, t, d;

	cin >> V >> E;

	vector<vector<int> > M(V, vector<int>(V, 1 << 27));

	for (int i = 0; i < E; i++)
	{
		cin >> s >> t >> d;

		M[s][t] = d;
	}

	vector<vector<int> > dp(V, vector<int>(1 << V, 1 << 27));

	dp[0][0] = 0;

	for (int i = 0; i < V; i++)
	{
		for (int j = 0; j < V; j++)
		{
			for (int k = 0; k < (1 << V); k++)
			{
				for (int l = 0; l < V; l++)
				{
					if ((k / (1 << j)) % 2 == 1)
					{
						dp[j][k] = min(dp[j][k], dp[l][k - (1 << j)] + M[l][j]);
					}
				}
			}
		}
	}

	cout << (dp[0][(1 << V) - 1] != (1 << 27) ? dp[0][(1 << V) - 1] : -1) << endl;

	return 0;
}

// Time : O ( V ^ 3 * 2 ^ V ) = 110,592,000
```