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

dist = [[float('inf') for _ in range(V)] for _ in range(V)]
for _ in range(E):
    s, t, d = map(int, input().split())
    dist[s][t] = d

# 2**V回、-1がV個の配列を作る
dp = [[-1]*V for _ in range(1<<V)]

def rec(s, v, dp):
    """
    s: 訪れた集合のbit演算（すべて訪れたら2**V-1になる）
    v: いまいる点
    dp: dp配列
    """
    if dp[s][v] >= 0:
        return dp[s][v]
    
    # すべて訪れており、かつ今いる点が0（始点）の場合
    if s == (1<<V)-1 and v == 0:
        dp[s][v] = 0
        return 0

    res = float('inf')
    for u in range(V):
    # まだuに訪れていないときuに訪れる。コストを再帰的に出す
        if (s>>u&1) == 0:
            res = min(res, rec(s|(1<<u), u, dp) + dist[v][u])

    dp[s][v] = res
    return res

ans = rec(0, 0, dp)

if ans == float('inf'):
    print(-1)
else:
    print(ans)
