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
import sys
input = sys.stdin.readline

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

#dist[i][j]: i -> jへの距離
dist = [[float("inf")] * V for i in range(V)]
for _ in range(E):
    s, t, d = map(int, input().split())
    dist[s][t] = d

dp = [[-1] * V for i in range(2**V)]

#dp[s][v]: 集合sを経由して今vに居るとき、始点に戻る最短経路長
def rec(s, v, dp):
    #　計算済みかどうか
    if dp[s][v] >= 0:
        return dp[s][v]
    
    #　すべての頂点を訪れたか
    if s == 2**V - 1 and v == 0:
        dp[s][v] = 0
        return 0
    
    res = 10**100
    for i in range(V):
        if (s>>i & 1) == 0:
            # i の箇所が訪れていないとき（0の時）次はiに行く
            # s|(1<<i) は、今まで訪れたsに加えて、頂点iをおとずれたことを加えた和集合
            res = min(res, rec(s|(1<<i), i, dp) + dist[v][i])
    
    dp[s][v] = res
    return dp[s][v]

ans = rec(0, 0, dp)
if ans == 10**100:
    ans = -1
print(ans)
