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
input = sys.stdin.readline
sys.setrecursionlimit(10**7)

V, E = map(int, input().split())
INF = 10**9
G = [[] for _ in range(V)]
for _ in range(E):
    s, t, d = map(int, input().split())
    G[t].append((s, d))
def rec(bit, v):
    if dp[bit][v] != INF: return dp[bit][v]
    before = bit^(1<<v)
    for u, d in G[v]:
        if before&(1<<u):
            dp[bit][v] = min(dp[bit][v], rec(before, u)+d)
        if before == 0 and u == 0:
            dp[bit][v] = min(dp[bit][v], d)
    return dp[bit][v]
# dp[S][v]: 集合Sの中でvが最後になる時の最短距離
dp = [[INF]*V for _ in range(1<<V)]
ans = INF
ans = min(ans, rec((1<<V)-1, 0))
if ans != INF:
    print(ans)
else:
    print(-1)
