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, raw_input().split())
g = [[] for i in xrange(V)]
for i in xrange(E):
    s, t, d = map(int, raw_input().split())
    g[s].append((t, d))
INF = 10**9
dp = [[-1]*V for i in xrange(2**V)]
tmp = {}
def dfs(state, v, s):
    if dp[state][v] != -1:
        return dp[state][v]
    if state == 2**V-1:
        for t, d in g[v]:
            if t==s:
                dp[state][v] = d
                return d
        return INF
    res = INF
    idx = -1
    for t, d in g[v]:
        if state & (1 << t):
            continue
        r = dfs(state | (1 << t), t, s) + d
        if r < res:
            idx = t
        res = min(res, r)

    tmp[state, v] = (idx, res)

    return res
def solve(v):
    global dp
    dp = [[-1]*V for i in xrange(2**V)]
    return dfs(1<<v, v, v)
res = min(solve(v) for v in xrange(V))
print res if res<INF else -1