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
# http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=DPL_2_A&lang=ja

import sys
sys.setrecursionlimit(10 ** 9)

V, E = map(int, input().split())

cost = [[float("inf")]*V for _ in range(V)]
for _ in range(E):
    s, t, d = map(int, input().split())
    cost[s][t] = d

dp = [[float("inf")]*(V) for _ in range((1<<V))]

dp[0][0] = 0                                # どこにも訪れていない状態で、頂点0を出発点とする。
                                            # 最後に頂点0に戻ってくるために頂点0にフラグは立てないようにしている。

for S in range(1<<V):
    for v in range(V):
        if not (1<<v)&S:                    # vをすでに訪れている場合を除く
            dp[S|(1<<v)][v] = min(dp[S][u] + cost[u][v] for u in range(V))
                                            # "if (1<<u)&S" が必要に見えるが、Sにuが含まれないようなdpテーブルは、
                                            # dp[0][0]を除いて全てinfから動かない（漸化式の左辺に現れない）ので、自動的に除かれる。
                                            # 逆に、"if (1<<u)&S"を入れようとすると、初期化を"dp[1][0]=0"として、
                                            # かつ最後に頂点0に戻る場合を別に処理する必要があるので少し面倒。

if dp[(1<<V)-1][0] == float("inf"):
    print(-1)
else:
    print(dp[(1<<V)-1][0])


