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
def tsp(matrix):
    INF = float("inf")
    n = len(matrix)

    # dp[bit_state][i] := すでにbit_stateの状態を訪れていて、
    #                     最後に訪問したのがi番目のときの最小値
    dp = [[INF]*n for i in range(1 << n)]
    dp[1 << 0][0] = 0
 
    for bit_state in range(1 << n):
        for i in range(n):
            # bit_stateにiを含んでいる
            if (bit_state >> i) & 1:
                for j in range(n):
                    # bit_stateにjを含んでいない
                    if not ((bit_state >> j) & 1):
                        dp[bit_state|(1 << j)][j] = min(dp[bit_state|(1 << j)][j], dp[bit_state][i] + matrix[i][j])
    ans = INF
    for i in range(n):
        ans = min(dp[(1 << n) - 1][i] + matrix[i][0], ans)
    return ans

n, m = map(int, input().split())
info = [list(map(int, input().split())) for i in range(m)]
INF = float("inf")

matrix = [[INF]*n for i in range(n)]
for i in range(m):
    a, b, cost = info[i]
    matrix[a][b] = cost
for i in range(n):
    matrix[i][i] = 0

ans = tsp(matrix)
if ans == INF:
    print(-1)
else:
    print(ans)
