#include <bits/stdc++.h>
 
using namespace std;

using usize = ::std::size_t;
//using u64 = ::std::uint_least64_t;
typedef int u64;
//static constexpr u64 Inf = ::std::numeric_limits<u64>::max() / 2;
const int Inf = 0x3f3f3f3f;

u64 N, V, E;
u64 dp[1 << 16][16], g[16][16];

u64 dfs(int mask , int u){
    
    
    if (mask == (1 << V) - 1) {
        return g[u][0];
    }
    
    if (~dp[mask][u]) return dp[mask][u];
    
    u64 r = Inf;
    
    for (int v = 0; v < V; v++) {
        if (v != u && !(mask & (1 << v))) {
            r = min(r, g[u][v] + dfs(mask | (1 << v), v));
        }
    }
    
    return dp[mask][u] = r;
    
}

int main(int argc, char *argv[])
{
    
 
    cin.tie(0);
    ios::sync_with_stdio(false);
    
    cin >> V >> E;
    memset(g, Inf, sizeof(g));
    for (int i = 0; i < E; i++) {
        u64 v1, v2;
        cin >> v1 >> v2;
        cin >> g[v1][v2];
    }
    memset(dp, -1, sizeof(dp));
    
    dfs(1, 0);
    
    if (dp[1][0] >= Inf) {
        cout << -1 << endl;
    } else {
        cout << dp[1][0] << endl;
    }
    
    
    return 0;
}


### Prompt

Please formulate 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;

using usize = ::std::size_t;
//using u64 = ::std::uint_least64_t;
typedef int u64;
//static constexpr u64 Inf = ::std::numeric_limits<u64>::max() / 2;
const int Inf = 0x3f3f3f3f;

u64 N, V, E;
u64 dp[1 << 16][16], g[16][16];

u64 dfs(int mask , int u){
    
    
    if (mask == (1 << V) - 1) {
        return g[u][0];
    }
    
    if (~dp[mask][u]) return dp[mask][u];
    
    u64 r = Inf;
    
    for (int v = 0; v < V; v++) {
        if (v != u && !(mask & (1 << v))) {
            r = min(r, g[u][v] + dfs(mask | (1 << v), v));
        }
    }
    
    return dp[mask][u] = r;
    
}

int main(int argc, char *argv[])
{
    
 
    cin.tie(0);
    ios::sync_with_stdio(false);
    
    cin >> V >> E;
    memset(g, Inf, sizeof(g));
    for (int i = 0; i < E; i++) {
        u64 v1, v2;
        cin >> v1 >> v2;
        cin >> g[v1][v2];
    }
    memset(dp, -1, sizeof(dp));
    
    dfs(1, 0);
    
    if (dp[1][0] >= Inf) {
        cout << -1 << endl;
    } else {
        cout << dp[1][0] << endl;
    }
    
    
    return 0;
}


```