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


def main():
    V, E = map(int, input().split())
    INF = float("inf")
    G = [[-1 for _ in range(V)] for _ in range(V)]

    for i in range(E):
        s,t,d = map(int, input().split())
        G[s][t] = d

    # dp[bit][i] bitの１が立っている場所を訪問済みで、現在iにいる
    dp = [[INF for _ in range(V)] for _ in range(1 << V)]
    # 各所一回だけ訪問して一周できるなら、どこから始めても同じなので0のみ初期化
    dp[0][0] = 0



    for bit in range(1 << V):
        for curr in range(V):
            for nxt in range(V):
                # 現在地と次の場所が同じなら飛ばす
                if curr == nxt:
                    continue
                # 次行くところが既に訪問済みなら飛ばす
                if bit & (1 << nxt):
                    continue
                # 現在値から次の目的地にいけなければ飛ばす
                if G[curr][nxt] == -1:
                    continue
                dp[bit | (1 << nxt)][nxt] = min(dp[bit | (1 << nxt)][nxt], dp[bit][curr] + G[curr][nxt])

    ans = dp[-1][0]
    print(ans if ans != INF else -1)
        
if __name__ == "__main__":
    main()
