#include <cstdio>
#include <algorithm>

using namespace std;

const int V_MAX = 15;
const int INF = 10000000;

int n, e;
int g[V_MAX][V_MAX];
int dp[1 << V_MAX][V_MAX];

int tsp(int visited, int v)
{
    if (dp[visited][v] != -1)
        return dp[visited][v];

    if (visited == (1 << n) - 1 && v == 0)
        return 0;

    int res = INF;
    for (int u = 0; u < n; ++u) {
        if (!(visited & (1 << u)) && g[v][u] != INF) {
            res = min(res, tsp(visited | (1 << u), u) + g[v][u]);
        }
    }
    return dp[visited][v] = res;
}

int main()
{
    scanf("%d %d", &n, &e);
    fill(&g[0][0], &g[n][0], INF);
    for (int i = 0; i < e; ++i) {
        int s, t, d;
        scanf("%d %d %d", &s, &t, &d);
        g[s][t] = d;
    }

    fill(&dp[0][0], &dp[1 << V_MAX][0], -1);
    int d = tsp(0, 0);
    if (d == INF)
        printf("-1\n");
    else
        printf("%d\n", d);

    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 <cstdio>
#include <algorithm>

using namespace std;

const int V_MAX = 15;
const int INF = 10000000;

int n, e;
int g[V_MAX][V_MAX];
int dp[1 << V_MAX][V_MAX];

int tsp(int visited, int v)
{
    if (dp[visited][v] != -1)
        return dp[visited][v];

    if (visited == (1 << n) - 1 && v == 0)
        return 0;

    int res = INF;
    for (int u = 0; u < n; ++u) {
        if (!(visited & (1 << u)) && g[v][u] != INF) {
            res = min(res, tsp(visited | (1 << u), u) + g[v][u]);
        }
    }
    return dp[visited][v] = res;
}

int main()
{
    scanf("%d %d", &n, &e);
    fill(&g[0][0], &g[n][0], INF);
    for (int i = 0; i < e; ++i) {
        int s, t, d;
        scanf("%d %d %d", &s, &t, &d);
        g[s][t] = d;
    }

    fill(&dp[0][0], &dp[1 << V_MAX][0], -1);
    int d = tsp(0, 0);
    if (d == INF)
        printf("-1\n");
    else
        printf("%d\n", d);

    return 0;
}
```