#include <bits/stdc++.h>
#pragma GCC optimize("O2")
#define all(c) c.begin(), c.end()
#define pb push_back
#define ll long long
const int MAXN = 1e5 + 5;

using namespace std;

int dp[15][1<<15], n;
vector<pair<int,int>> ve[15];

int rec(int u, int mask) {
  if (mask == (1<<n)-1) return 0;
  int &ret = dp[u][mask];
  if (ret != -1) return ret;
  ret = 1e9;
  for (auto v : ve[u])
    if (!(mask&(1<<v.first)) && (v.first != 0 || mask == (1<<n)-2))
      ret = min(ret, v.second+rec(v.first, mask|(1<<v.first)));
  return ret;
}

int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int e; cin >> n >> e;
  while (e--) {
    int u, v, w;
    cin >> u >> v >> w;
    ve[u].pb({v, w});
  }
  memset(dp, -1, sizeof dp);
  int mini = rec(0, 0);
  cout << (mini == 1e9? -1 : mini) << "\n";
  return 0;
}

### Prompt

Your challenge is to write 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>
#pragma GCC optimize("O2")
#define all(c) c.begin(), c.end()
#define pb push_back
#define ll long long
const int MAXN = 1e5 + 5;

using namespace std;

int dp[15][1<<15], n;
vector<pair<int,int>> ve[15];

int rec(int u, int mask) {
  if (mask == (1<<n)-1) return 0;
  int &ret = dp[u][mask];
  if (ret != -1) return ret;
  ret = 1e9;
  for (auto v : ve[u])
    if (!(mask&(1<<v.first)) && (v.first != 0 || mask == (1<<n)-2))
      ret = min(ret, v.second+rec(v.first, mask|(1<<v.first)));
  return ret;
}

int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int e; cin >> n >> e;
  while (e--) {
    int u, v, w;
    cin >> u >> v >> w;
    ve[u].pb({v, w});
  }
  memset(dp, -1, sizeof dp);
  int mini = rec(0, 0);
  cout << (mini == 1e9? -1 : mini) << "\n";
  return 0;
}

```