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
import sys

V, E = map(int, sys.stdin.readline().strip().split())

# edges[i][j]:i→jへの距離
edges = [[float("inf")]*V for i in range(V)]
for i in range(E):
   s, t, d = map(int, sys.stdin.readline().strip().split())
   edges[s][t] = d

dp = [[-1] * V for i in range(1<<V)]

# 訪れた集合がs、今いる点がvの時に、v=0に戻る最短経路
def rec(bit, v):
    if dp[bit][v] >= 0:
        return dp[bit][v]

    if bit == (1<<V)-1 and v == 0:
        #全ての頂点を訪れた(bit = 11...11 and v = 0)
        dp[bit][v] = 0
        return 0

    res = float("inf")
    for u in range(V):
        if not (bit>>u & 1):
            #uに訪れていない時(uの箇所が0の時),次はuとすると
            res = min(res,rec(bit|(1<<u), u) + edges[v][u])

    dp[bit][v] = res
    return res

res = rec(0, 0)
print(-1 if res == float("inf") else res)
