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
#!/usr/bin/env python3

from pprint import pprint
from collections import deque, defaultdict
import itertools
import math
import sys

sys.setrecursionlimit(10 ** 6)
input = sys.stdin.buffer.readline
INF = 10 ** 10

n_nodes, m_edges = map(int, input().split())
dist = [[INF] * n_nodes for _ in range(n_nodes)]

for _ in range(m_edges):
    s, t, d = map(int, input().split())
    dist[s][t] = d

memory = {}
end = 2 ** n_nodes - 1


def rec(state, pos):
    if (state, pos) in memory:
        return memory[(state, pos)]

    if state == end:
        return dist[pos][0]

    mask = 1
    res = INF
    for pos_next in range(n_nodes):
        if mask & state:
            mask <<= 1
            continue

        res = min(res, rec(state | mask, pos_next) + dist[pos][pos_next])
        mask <<= 1

    memory[(state, pos)] = res
    return res


res = rec(1, 0)
if res == INF:
    print(-1)
else:
    print(res)

