#include "bits/stdc++.h"

#define INF 20000000
#define FER 0.0000000001

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<ll, int> pli;
typedef pair<double, double> pdd;

int dp[1 << 15][15];
int dist[15][15];

int main()
{
	cin.tie(0);
	ios::sync_with_stdio(false);

	for (int i = 0; i < 15; i++)for (int j = 0; j < 15; j++)
	{
		dist[i][j] = INF;
	}

	int V, E;
	cin >> V >> E;
	for (int i = 0; i < E; i++)
	{
		int s, t, d;
		cin >> s >> t >> d;
		dist[s][t] = d;
	}

	for (int i = 0; i < 1 << V; i++)
	{
		fill(dp[i], dp[i] + V, INF);
	}
	dp[(1 << V) - 1][0] = 0;

	for (int S = (1 << V) - 2; S >= 0; S--)
	{
		for (int v = 0; v < V; v++)
		{
			for (int u = 0; u < V; u++)
			{
				if (!(S >> u & 1))
				{
					dp[S][v] = min(dp[S][v], dp[S | (1 << u)][u] + dist[v][u]);
				}
			}
		}
	}

	printf(dp[0][0] == INF ? "-1\n" : "%d\n", dp[0][0]);

	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"

#define INF 20000000
#define FER 0.0000000001

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<ll, int> pli;
typedef pair<double, double> pdd;

int dp[1 << 15][15];
int dist[15][15];

int main()
{
	cin.tie(0);
	ios::sync_with_stdio(false);

	for (int i = 0; i < 15; i++)for (int j = 0; j < 15; j++)
	{
		dist[i][j] = INF;
	}

	int V, E;
	cin >> V >> E;
	for (int i = 0; i < E; i++)
	{
		int s, t, d;
		cin >> s >> t >> d;
		dist[s][t] = d;
	}

	for (int i = 0; i < 1 << V; i++)
	{
		fill(dp[i], dp[i] + V, INF);
	}
	dp[(1 << V) - 1][0] = 0;

	for (int S = (1 << V) - 2; S >= 0; S--)
	{
		for (int v = 0; v < V; v++)
		{
			for (int u = 0; u < V; u++)
			{
				if (!(S >> u & 1))
				{
					dp[S][v] = min(dp[S][v], dp[S | (1 << u)][u] + dist[v][u]);
				}
			}
		}
	}

	printf(dp[0][0] == INF ? "-1\n" : "%d\n", dp[0][0]);

	return 0;
}

```