#include <bits/stdc++.h>
using namespace std;
const int INF = (1 << 29);

int V, E;
vector<vector<int>> dist;
vector<vector<int>> dp;

int rec(int bit, int v)
{
  // 計算済みのとき
  if (~dp.at(bit).at(v))
  {
    return dp.at(bit).at(v);
  }
  // 初期条件
  if (bit == 0)
  {
    if (v == 0)
    {
      return dp.at(bit).at(v) = 0;
    }
    else
    {
      return dp.at(bit).at(v) = INF;
    }
  }
  // 再帰ステップ
  int res = INF;
  // 最短経路における、ひとつ前の頂点がu
  for (int u = 0; u < V; u++)
  {
    if (!(bit & (1 << u)))
    {
      continue;
    }
    if (dist.at(u).at(v) == INF)
    {
      continue;
    }
    int prev_bit = bit & ~(1 << u);
    res = min(res, rec(prev_bit, u) + dist.at(u).at(v));
  }
  return dp.at(bit).at(v) = res;
}

int main()
{
  cin >> V >> E;
  dist.resize(V, vector<int>(V, INF));
  for (int i = 0; i < E; i++)
  {
    int s, t, d;
    cin >> s >> t >> d;
    dist.at(s).at(t) = d;
  }
  dp.resize((1 << V), vector<int>(V, -1));
  int res = rec((1 << V) - 1, 0);
  if (res == INF)
  {
    res = -1;
  }
  cout << res << endl;
}

### Prompt

Develop a solution in cpp 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;
const int INF = (1 << 29);

int V, E;
vector<vector<int>> dist;
vector<vector<int>> dp;

int rec(int bit, int v)
{
  // 計算済みのとき
  if (~dp.at(bit).at(v))
  {
    return dp.at(bit).at(v);
  }
  // 初期条件
  if (bit == 0)
  {
    if (v == 0)
    {
      return dp.at(bit).at(v) = 0;
    }
    else
    {
      return dp.at(bit).at(v) = INF;
    }
  }
  // 再帰ステップ
  int res = INF;
  // 最短経路における、ひとつ前の頂点がu
  for (int u = 0; u < V; u++)
  {
    if (!(bit & (1 << u)))
    {
      continue;
    }
    if (dist.at(u).at(v) == INF)
    {
      continue;
    }
    int prev_bit = bit & ~(1 << u);
    res = min(res, rec(prev_bit, u) + dist.at(u).at(v));
  }
  return dp.at(bit).at(v) = res;
}

int main()
{
  cin >> V >> E;
  dist.resize(V, vector<int>(V, INF));
  for (int i = 0; i < E; i++)
  {
    int s, t, d;
    cin >> s >> t >> d;
    dist.at(s).at(t) = d;
  }
  dp.resize((1 << V), vector<int>(V, -1));
  int res = rec((1 << V) - 1, 0);
  if (res == INF)
  {
    res = -1;
  }
  cout << res << endl;
}

```