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

n,w = map(int,input().split())

#d[i][j]:i→jへの距離
d = [[float("inf")]*n for i in range(n)]
for i in range(w):
   x,y,z = map(int,input().split())
   d[x][y] = z 

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

#訪れた集合がs、今いる点がvの時０に戻る最短経路
def rec(s,v,dp):
    if dp[s][v] >= 0:
        return dp[s][v]
        
    if s == (1<<n)-1 and v == 0:
        #全ての頂点を訪れた(s = 11...11 and v = 0)
        dp[s][v] = 0
        return 0
    
    res = float("inf")
    for u in range(n):
        if (s>>u&1) == 0:
            #uに訪れていない時(uの箇所が0の時),次はuとすると
#            print("s:",bin(s),"u:",u,"v:",v,"sp:",bin(s|(1<<u)),d[v][u])
            res = min(res,rec(s|(1<<u),u,dp)+d[v][u])
            
    dp[s][v] = res
    return res

rec1=rec(0,0,dp)
if rec1==float("inf"):
    print(-1)
else:
    print(rec1)

