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

file_input = sys.stdin

V, E = map(int, file_input.readline().split())

no_route = 15001

# (v-1) th vertex is excluded
dp = [[no_route] * (V - 1) for i in range(1 << (V - 1))]

adj = [[] for i in range(V)]

for line in file_input:
    s, t, d = map(int, line.split())
    if s == V - 1:
        # Record the distance from the (v-1) th vertex to the adjacent vertex
        dp[1 << t][t] = d
    else: 
        # Store distance from source, not target
        adj[t].append((s, d))

# DP
for b in range(1, 1 << (V - 1)):
    for i in range(V - 1):
        b_i = 1 << i
        if b & b_i:
            pre_bit = b ^ b_i
            dp_pre = dp[pre_bit]
            for s, d in adj[i]:
                pre_rec = dp_pre[s]
                if pre_rec != no_route:
                    dp[b][i] = min(dp[b][i], pre_rec + d)

ans = no_route

# A state where all vertices other than (v-1) have been visited.
dp_comp = dp[(1 << (V - 1)) - 1]

for s, d in adj[V - 1]:
    comp_rec = dp_comp[s]
    if comp_rec != no_route:
        ans = min(ans, comp_rec + d)

if ans == no_route:
    print(-1)
else:
    print(ans)