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 solve():
    from heapq import heappop, heappush
    V, E = map(int, input().split())
    edges = [[] for _ in [0]*V]
    for _ in [0]*E:
        s, t, d = map(int, input().split())
        edges[s].append((t, d))

    result = float("inf")
    for i in range(V):
        q = [(0, i, {i})]
        while q:
            cost, v, visited = heappop(q)
            if len(visited) < V:
                for dest, d_cost in edges[v]:
                    if dest not in visited:
                        heappush(q, (cost+d_cost, dest, visited | {dest}))
            elif len(visited) == V:
                for dest, d_cost in edges[v]:
                    if dest == i:
                        heappush(q, (cost+d_cost, i, visited | {-1}))
            else:
                if result > cost:
                    result = cost
                break

    print(result if result < float("inf") else -1)


if __name__ == "__main__":
    solve()