#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)

typedef long long ll;
typedef pair<int, int> P;

const ll mod =  1000000007;  //出力は (ans % mod + mod) % mod  (負の剰余を正にする)
const int inf = 1e9;
const long long INF = 1LL << 60;   // INF = 1152921504606846976

int main()
{
    int V,e; cin >> V >> e;
    ll dp[(1<<V)][V];
    rep(i,(1<<V)) rep(j,V) dp[i][j] = INF;
    dp[0][0] = 0;
    ll g[V][V];
    rep(i,V) rep(j,V) g[i][j] = INF;
    rep(i,e){
        int s,t,d; cin >> s >> t >> d;
        g[s][t] = d;
    }
    for(int S = 0; S < (1 << V); S++){
        for(int v = 0; v < V; v++){
            for(int u = 0; u < V; u++){
                if(S & (1<<v)){
                    //もらうDP
                    //S-{v}からSになるときの最小値を更新していく
                    dp[S][v] = min(dp[S][v],dp[S - (1<<v)][u] + g[u][v]);
                }
            }
        }
    }
    if(dp[(1<<V)-1][0] != INF){
        cout << dp[(1<<V)-1][0] << endl;
    }else{
        cout << -1 << endl;
    }
}

### 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>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)

typedef long long ll;
typedef pair<int, int> P;

const ll mod =  1000000007;  //出力は (ans % mod + mod) % mod  (負の剰余を正にする)
const int inf = 1e9;
const long long INF = 1LL << 60;   // INF = 1152921504606846976

int main()
{
    int V,e; cin >> V >> e;
    ll dp[(1<<V)][V];
    rep(i,(1<<V)) rep(j,V) dp[i][j] = INF;
    dp[0][0] = 0;
    ll g[V][V];
    rep(i,V) rep(j,V) g[i][j] = INF;
    rep(i,e){
        int s,t,d; cin >> s >> t >> d;
        g[s][t] = d;
    }
    for(int S = 0; S < (1 << V); S++){
        for(int v = 0; v < V; v++){
            for(int u = 0; u < V; u++){
                if(S & (1<<v)){
                    //もらうDP
                    //S-{v}からSになるときの最小値を更新していく
                    dp[S][v] = min(dp[S][v],dp[S - (1<<v)][u] + g[u][v]);
                }
            }
        }
    }
    if(dp[(1<<V)-1][0] != INF){
        cout << dp[(1<<V)-1][0] << endl;
    }else{
        cout << -1 << endl;
    }
}

```