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
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int INF = 1 << 30;
const int MOD = 1e9 + 7;
int n, m;
struct edge {
  int to, cost;
  edge(int to, int cost) : to(to), cost(cost) {}
};
vector<edge> G[18];
int dp[1 << 18][18];
int dfs(int now, int bit, const int start) {
  if (__builtin_popcount(bit) == n) {
    for (int u = 0; u < G[now].size(); u++)
      if (G[now][u].to == start) return G[now][u].cost;
    return INF;
  }
  if (dp[bit][now]) return dp[bit][now];
  int res = INF;
  for (int u = 0; u < G[now].size(); u++) {
    edge e = G[now][u];
    if (bit & (1 << e.to)) continue;
    res = min(res, dfs(e.to, bit | (1 << e.to), start) + e.cost);
  }
  return dp[bit][now] = res;
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  memset(dp, 0, sizeof(dp));
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int s, t, d;
    cin >> s >> t >> d;
    G[s].push_back(edge(t, d));
  }
  int ans = INF;
  for (int i = 0; i < n; i++) ans = min(ans, dfs(i, 0 | (1 << i), i));
  cout << (ans == INF ? -1 : ans) << endl;
  return 0;
}
