#include <bits/stdc++.h>
using namespace std;
using i64 = int64_t;
using vi = vector<i64>;
using vvi = vector<vi>;

int main() {
    int v, e;
    cin >> v >> e;
    using ii = pair<int, i64>;
    vector<vector<ii>> adj(v);
    for (int i = 0; i < e; i++) {
        int s, t, d;
        cin >> s >> t >> d;
        adj[s].push_back(ii(t, d));
    }

    vvi dp(1 << v, vi(v, 1e18));
    // dp[i][j]: すでにiを回って今jにいる
    dp[0][0] = 0;
    for (int i = 0; i < 1 << v; i++) {
        for (int j = 0; j < v; j++) {
            for (auto &p: adj[j]) {
                if ((i >> p.first) & 1) continue;
                dp[i | (1 << p.first)][p.first] = min(dp[i | (1 << p.first)][p.first], dp[i][j] + p.second);
            }   
        }
    }
    cout << (dp[(1 << v) - 1][0] == 1e18 ? -1 : dp[(1 << v) - 1][0]) << endl;
}

### 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;
using i64 = int64_t;
using vi = vector<i64>;
using vvi = vector<vi>;

int main() {
    int v, e;
    cin >> v >> e;
    using ii = pair<int, i64>;
    vector<vector<ii>> adj(v);
    for (int i = 0; i < e; i++) {
        int s, t, d;
        cin >> s >> t >> d;
        adj[s].push_back(ii(t, d));
    }

    vvi dp(1 << v, vi(v, 1e18));
    // dp[i][j]: すでにiを回って今jにいる
    dp[0][0] = 0;
    for (int i = 0; i < 1 << v; i++) {
        for (int j = 0; j < v; j++) {
            for (auto &p: adj[j]) {
                if ((i >> p.first) & 1) continue;
                dp[i | (1 << p.first)][p.first] = min(dp[i | (1 << p.first)][p.first], dp[i][j] + p.second);
            }   
        }
    }
    cout << (dp[(1 << v) - 1][0] == 1e18 ? -1 : dp[(1 << v) - 1][0]) << endl;
}

```