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
from collections import defaultdict,deque
import sys,heapq,bisect,math,itertools,string,queue,copy,time
sys.setrecursionlimit(10**8)
INF = float('inf')
mod = 10**9+7
eps = 10**-7
def inp(): return int(input())
def inpl(): return list(map(int, input().split()))
def inpl_str(): return list(input().split())

N,M = inpl()
lines = defaultdict(set)
for _ in range(M):
    s,t,d = inpl()
    lines[s].add((t,d))


S = 0
dp = [[INF]*N for _ in range(1<<(N+1)-1)]
dp[0][S] = 0
for bit in range(1<<(N+1)-1):
    for s in range(N):
        now = dp[bit][s]
        for t,c in lines[s]:
            if bit & (1<<t): # 訪れたことがある
                continue
            else:
                dp[bit+(1<<t)][t] = min(dp[bit+(1<<t)][t],now+c)

ans = dp[-1][S]
if ans == INF:
    print(-1)
else:
    print(dp[-1][S])

