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
#よくわからないがAC

import sys
sys.setrecursionlimit(10**7)

n,m=map(int,input().split())
#G[i]=iに来れるノード
G=[[] for i in range(n)]
for i in range(m):
    s,t,d=map(int,input().split())
    G[t].append((s,d))


dp=[[float('inf')]*n for i in range(2**n)]

#頂点0からとする
dp[0][0]=0
#0からbitを辿ってvに来る(vはSに入っている)
def rec(bit,v):
    if dp[bit][v]!=float('inf'):
        return dp[bit][v]
    
    res=float('inf')
    #Sからvを取る
    prev_bit=bit&~(1<<v)
    #print(bit,v,prev_bit)
    for u,cost in G[v]:
        #下手
        if prev_bit==0 and u==0:
            res=cost
            break
        if not (prev_bit&(1<<u)):
            continue
        
        res=min(res,rec(prev_bit,u)+cost)
    
    dp[bit][v]=res
    return dp[bit][v]

ans=rec(2**n-1,0)
if ans==float('inf'):
    print(-1)
else:
    print(ans)




