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
# テーブル埋め
INF = float('inf')
V, E = [int(x) for x in input().split()]
D = [[INF for _ in range(V)] for _ in range(V)]
for i in range(V):
    D[i][i] = 0
for j in range(E):
    s, t, d = [int(x) for x in input().split()]
    D[s][t] = d

dp = [[INF for _ in range(V)] for _ in range(1 << V)]
for S in range((1 << V) - 1, -1, -1):
    for v in range(V):
        if S == (1 << V) - 1:
            if v == 0:
                dp[S][v] = 0
        else:
            # 次に行くuを決める
            for u in range(V):
                if (S >> u) & 1 == 0:
                    dp[S][v] = min(dp[S][v], D[v][u] + dp[S + (1 << u)][u])

if dp[0][0] == INF:
    print(-1)
else:
    print(dp[0][0])
