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
sys.setrecursionlimit(10 ** 7)
read = sys.stdin.buffer.read
inp = sys.stdin.buffer.readline
def inpS(): return inp().rstrip().decode()
readlines = sys.stdin.buffer.readlines
MOD = 10 ** 9 + 7
INF = 1 << 60

# ---------------------------------------------------
# ---------------------------------------------------
# ---------------------------------------------------

def resolve():
    def rec(S, v):
        if dp[S][v] >= 0:
            return dp[S][v]
        if S == (1<<V)-1 and v==0:
            dp[S][v] = 0
            return 0
        res = INF
        for to in range(V):
            if not (S>>to & 1):
                res = min(res, rec((S | (1<<to)), to) + dist[v][to])

        dp[S][v] = res
        return res

    V, E = map(int, input().split())
    dist = [[INF] * V for _ in range(V)]
    for i in range(E):
        x, y, d = map(int, input().split())
        dist[x][y] = d

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

    res = rec(0, 0)
    if res == INF:
        print(-1)
    else:
        print(res)

if __name__ == '__main__':
    resolve()
