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
# ハミルトン閉路はどこから初めても答えは変わらない
V, E = map(int, input().split())
INF = 10**18
dist = [[INF] * V for _ in range(V)]
for _ in range(E):
  s, t, d = map(int, input().split())
  dist[s][t] = d
  
# dp[i][j]:= 状態i(2**V)にするための距離の最小値
# 最後に訪れたのがj
dp = [[INF] * V for _ in range(2**V)]

# 始点を0とする
dp[0][0] = 0

for s in range(2**V):# 現在の状態
  for j in range(V):# 次の頂点
    if (s >> j & 1)==0:# jが集合Sに含まれない
      for i in range(V):# 今いる頂点
        if i != j:
          dp[s|1<<j][j] = min(dp[s|1<<j][j],dp[s][i]+dist[i][j])

print(-1) if dp[2**V-1][0] == INF else print(dp[2**V-1][0])
