#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<ll> VL;
typedef vector<VL> VVL;
typedef pair<int, int> PII;

#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()
#define MOD 1000000007
#define INF (1LL<<30)
#define LLINF (1LL<<60)
#define PI 3.14159265359
#define EPS 1e-12
#define int ll

int n, d[16][16], dp[1<<15][16];
signed main(void)
{
  int m;
  cin >> n >> m;
  REP(i, n) REP(j, n) d[i][j] = INF;
  REP(i, m) {
    int s, t, dd;
    cin >> s >> t >> dd;
    d[s][t] = dd;
  }
  REP(i, 1<<n) fill(dp[i], dp[i]+n, INF);
  dp[(1<<n)-1][0] = 0;

  for(int s=(1<<n)-2; s>=0; --s) {
    REP(v, n) REP(u, n) {
      if(!(s>>u&1)) dp[s][v] = min(dp[s][v], dp[s|1<<u][u] + d[v][u]);
    }
  }
  if(dp[0][0] >= INF) cout << -1 << endl;
  else cout << dp[0][0] << endl;
  return 0;
}
### Prompt

In cpp, your task is to solve 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 ll;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<ll> VL;
typedef vector<VL> VVL;
typedef pair<int, int> PII;

#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()
#define MOD 1000000007
#define INF (1LL<<30)
#define LLINF (1LL<<60)
#define PI 3.14159265359
#define EPS 1e-12
#define int ll

int n, d[16][16], dp[1<<15][16];
signed main(void)
{
  int m;
  cin >> n >> m;
  REP(i, n) REP(j, n) d[i][j] = INF;
  REP(i, m) {
    int s, t, dd;
    cin >> s >> t >> dd;
    d[s][t] = dd;
  }
  REP(i, 1<<n) fill(dp[i], dp[i]+n, INF);
  dp[(1<<n)-1][0] = 0;

  for(int s=(1<<n)-2; s>=0; --s) {
    REP(v, n) REP(u, n) {
      if(!(s>>u&1)) dp[s][v] = min(dp[s][v], dp[s|1<<u][u] + d[v][u]);
    }
  }
  if(dp[0][0] >= INF) cout << -1 << endl;
  else cout << dp[0][0] << endl;
  return 0;
}
```