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 TS(G,v,e):
    INF = 16000
    DP = [[INF for _ in range(v)] for _ in range(2**v)]
    #DP[i][j]:訪問済みエリアのリストがiでjにいるような最短距離
    DP[0][0] = 0
    #   どれも訪問してない状態は00...0(長さv)として訪問済みを表現
    for i in range(2**v):
        for j in range(v):
            for k in range(v):
                if i ^ 2**k < i and j != k \
                and DP[i ^ 2**k][k] + G[k][j] < DP[i][j]:
                    DP[i][j] = DP[i ^ 2**k][k] + G[k][j]

    if DP[2**v-1][0] != INF:
        return DP[2**v-1][0]
    else:
        return -1

def main():
    v,e = map(int,input().split())
    INF = 16000
    G = [[INF for _ in range(v)] for _ in range(v)]
    for i in range(e):
        s,t,d = map(int,input().split())
        G[s][t] = d

    print(TS(G,v,e))

if __name__ == "__main__":
    main()

