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
INF = 10**12
V, E = map(int, input().split())
d = [[INF]*(V) for _ in range(V)]
for i in range(E):
    s, t, dd = map(int, input().split())
    d[s][t] = dd
def f():
    #dp[i][j] : i　はすでに訪れた集合, 現在いる状態から残りの全ての頂点を巡って頂点０に帰るようなパスの重みの最小値
    dp = [[INF]*(V) for _ in range(1<<V)]
    dp[(1<<V)-1][0] = 0
    for i in range((1<<V)-2, -1, -1):
        for j in range(V):
            for k in range(V):
                if not (i>>k & 1):
                    dp[i][j] = min(dp[i][j], dp[i | 1 << k][k]+ d[j][k])
    if dp[0][0] < INF:
        print(dp[0][0])
    else:
        print(-1)
    
f()
