#include <bits/stdc++.h>

using namespace std;
typedef long long int ll;
typedef pair<int, int> P;
typedef pair<ll, ll> Pll;
typedef vector<int> Vi;
//typedef tuple<int, int, int> T;
#define FOR(i,s,x) for(int i=s;i<(int)(x);i++)
#define REP(i,x) FOR(i,0,x)
#define ALL(c) c.begin(), c.end()
#define DUMP( x ) cerr << #x << " = " << ( x ) << endl

const int dr[4] = {-1, 0, 1, 0};
const int dc[4] = {0, 1, 0, -1};

const int INF = (int)1e7;

int main() {
  // use scanf in CodeForces!
  cin.tie(0);
  ios_base::sync_with_stdio(false);

  int V, E; cin >> V >> E;
  vector<vector<int>> dp(1 << V, vector<int>(V, INF)), D(V, vector<int>(V, INF));

  REP(i, E) {
    int s, t, d; cin >> s >> t >> d;
    D[s][t] = d;
  }

  dp[0][0] = 0;
  for (int state = 0; state < (1 << V); state++) {
    REP(i, V) REP(j, V) if (not (state >> j & 1) and dp[state | (1 << j)][j] > dp[state][i] + D[i][j]) {
      dp[state | (1 << j)][j] = dp[state][i] + D[i][j];
    }
  }

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

using namespace std;
typedef long long int ll;
typedef pair<int, int> P;
typedef pair<ll, ll> Pll;
typedef vector<int> Vi;
//typedef tuple<int, int, int> T;
#define FOR(i,s,x) for(int i=s;i<(int)(x);i++)
#define REP(i,x) FOR(i,0,x)
#define ALL(c) c.begin(), c.end()
#define DUMP( x ) cerr << #x << " = " << ( x ) << endl

const int dr[4] = {-1, 0, 1, 0};
const int dc[4] = {0, 1, 0, -1};

const int INF = (int)1e7;

int main() {
  // use scanf in CodeForces!
  cin.tie(0);
  ios_base::sync_with_stdio(false);

  int V, E; cin >> V >> E;
  vector<vector<int>> dp(1 << V, vector<int>(V, INF)), D(V, vector<int>(V, INF));

  REP(i, E) {
    int s, t, d; cin >> s >> t >> d;
    D[s][t] = d;
  }

  dp[0][0] = 0;
  for (int state = 0; state < (1 << V); state++) {
    REP(i, V) REP(j, V) if (not (state >> j & 1) and dp[state | (1 << j)][j] > dp[state][i] + D[i][j]) {
      dp[state | (1 << j)][j] = dp[state][i] + D[i][j];
    }
  }

  cout << (dp[(1 << V) - 1][0] == INF ? -1 : dp[(1 << V) - 1][0]) << endl;
  return 0;
}
```