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 salesman(T, now_v, dp, graph, v):
    if dp[T][now_v] >= 0:
        return dp[T][now_v]
    if T == (1 << v) - 1 and now_v == 0:
        dp[T][now_v] = 0
        return dp[T][now_v]
    res = 10 ** 7
    for i in range(v):
        if not ((T >> i) & 1):
            res = min(res, salesman(T | (1 << i), i, dp, graph, v) + graph[now_v][i])
    dp[T][now_v] = res
    return dp[T][now_v]


def main():
    v, e = map(int, input().split())
    INF = 10 ** 7
    graph = [[INF for i in range(v)] for j in range(v)]
    for i in range(e):
        s, t, d = map(int, input().split())
        graph[s][t] = d
    dp = []
    for i in range(1 << v):
        dp.append([])
        for j in range(v):
            dp[i].append(-1)
    salesman(0, 0, dp, graph, v)
    if dp[0][0] >= INF:
        print(-1)
    else:
        print(dp[0][0])
    
    




if __name__ == '__main__':
    main()

