#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
static const int MAX_N = 16;
static const int INFTY = (1 << 29);
int V;
int G[MAX_N][MAX_N];
int dp[1 << MAX_N][MAX_N];

int rec(int S, int v) {
    if (dp[S][v] >= 0) return dp[S][v];
    if (S == (1 << V) -1 && v == 0) return dp[S][v] = 0;
    int res = INFTY;
    for (int u = 0;  u < V; u++) {
        if (! (S >> u & 1)) {
            res = min(res, rec(S | 1 << u, u) + G[v][u]);
        }
    }
    return dp[S][v] = res;
}


int main()
{
    int E, source, target, cost;
    scanf("%d %d", &V, &E);
    for (int i = 0; i < V; i++) fill(G[i], G[i]+V, INFTY);
    for (int i = 0; i < E; i++) {
        scanf("%d %d %d", &source, &target, &cost);
        G[source][target] = cost;
    }

    memset(dp, -1, sizeof (dp));
    int res = rec(0, 0);
    (res == INFTY) ? printf("%d\n", -1) : printf("%d\n", res);
    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 <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
static const int MAX_N = 16;
static const int INFTY = (1 << 29);
int V;
int G[MAX_N][MAX_N];
int dp[1 << MAX_N][MAX_N];

int rec(int S, int v) {
    if (dp[S][v] >= 0) return dp[S][v];
    if (S == (1 << V) -1 && v == 0) return dp[S][v] = 0;
    int res = INFTY;
    for (int u = 0;  u < V; u++) {
        if (! (S >> u & 1)) {
            res = min(res, rec(S | 1 << u, u) + G[v][u]);
        }
    }
    return dp[S][v] = res;
}


int main()
{
    int E, source, target, cost;
    scanf("%d %d", &V, &E);
    for (int i = 0; i < V; i++) fill(G[i], G[i]+V, INFTY);
    for (int i = 0; i < E; i++) {
        scanf("%d %d %d", &source, &target, &cost);
        G[source][target] = cost;
    }

    memset(dp, -1, sizeof (dp));
    int res = rec(0, 0);
    (res == INFTY) ? printf("%d\n", -1) : printf("%d\n", res);
    return 0;
}


```