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
#標準入力
V, E = map(int, input().split())
dist = [[float("inf")] * (V + 1) for i in range(V + 1)]
for i in range(E):
  s, t, d = map(int, input().split())
  dist[s][t] = d
  
for i in range(V):  
  dist[i][i] = 0
  
#print(dist)
#2 ** 15 = 1024 * 16 = 16384
dp = [[-1] * (V + 1) for i in range(2 ** V + 1)]

def rec(bit, v):
  #すでに探索済みだったらリターン
  if dp[bit][v] != -1:
    return dp[bit][v]
  
  #初期値
  if bit == (1 << 0):
    dp[bit][0] = 0
    return 0
  
  res = float("inf")
  prev_bit = bit & ~(1 << v)
  
  for i in range(V):
    if not (prev_bit & (1 << i)):
      continue
    else:
      res = min(res, rec(prev_bit, i) + dist[i][v])
  
  dp[bit][v] = res
  return res

ans = float("inf")
for i in range(V):
  ans = min(ans, rec((1 << V) - 1, i) + dist[i][0])
  #print(dp)

if ans >= 1000000:
  print(-1)
else:
  print(ans)
#print(dp)
