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
# http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=DPL_2_A&lang=ja

INF = 10**9
n,e = map(int, input().split())
gl = [ [INF]*(n) for _ in range(n)]

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

dp = [ [INF]*(n) for _ in range(2**n) ]
dp[1][0] = 0
for i in range(2**n):
    for j in range(n):
        if dp[i][j] == INF: continue
        for k in range(n):
            if (i>>k)%2 == 1: continue
            dp[i|(1<<k)][k] = min(dp[i][j]+gl[j][k], dp[i|(1<<k)][k])

ans = INF
for i in range(1,n): 
    if gl[i][0] != INF:
        ans = min(ans, dp[2**n-1][i] + gl[i][0])

if ans == INF: ans = -1
print(ans)
