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 math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools,queue

sys.setrecursionlimit(10**7)
INF = 20000
EPS = 1.0 / 10**10
MOD = 10**9+7

def LI(): return [int(x) for x in sys.stdin.readline().split()]

n, e = LI()
dist = [[INF for _ in range(n)] for _ in range(n)]
for _ in range(e):
    a,b,c = LI()
    dist[a][b] = c

dp = [[INF for _ in range(n)] for _ in range(1<<n)]

q = queue.Queue()

for i in range(1,n):
    if dist[0][i] < INF:
        dp[1 | (1<<i)][i] = dist[0][i]
        q.put((1 | (1<<i),i))

while not q.empty():
    bit, v = q.get()
    for u in range(1,n):
        if dist[v][u] < INF and not(bit & (1<<u)):
            next_bit = bit | (1<<u)
            if dp[next_bit][u]  == INF:
                q.put((next_bit, u))
            dp[next_bit][u] = min(dp[next_bit][u], dp[bit][v] + dist[v][u])

ans = INF
for i in range(1,n):
    ans = min(ans, dp[(1<<n)-1][i] + dist[i][0])
if ans < INF:
    print (ans)
else:
    print (-1)
