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
def main():
    inf=float("inf")
    v,e=map(int,input().split())
    con=[{} for _ in range(v)]
    for _ in range(e):
        s,t,d=map(int,input().split())
        con[s][t]=d

    #最終的に閉路が得られる場合、結果はスタート地点によらないので0→0を考えればよい
    #ij成分はiのbitで表現された頂点0を含まない部分集合のうち、j番目の頂点で終わるものの最短経路長
    dp=[[inf]*(v-1) for _ in range(2**(v-1))]
    #initialize
    for x in con[0]:
        dp[2**(x-1)][x-1]=con[0][x]
    
    for i in range(1,2**(v-1)):
        if bin(i).count("1")==1 : continue
        for j in range(v-1):
            if not i>>j &1:continue
            work=i-2**j
            mn=inf
            for k in range(v-1):
                if not work>>k &1 : continue
                if not j+1 in con[k+1] : continue
                if dp[work][k]==inf : continue
                if mn>dp[work][k]+con[k+1][j+1]:
                    mn=dp[work][k]+con[k+1][j+1]
            dp[i][j]=mn
    
    mn=inf
    #得られたグラフの終点と0をつなげられるか調べ最短距離を求める
    for i in range(v-1):
        if 0 in con[i+1]:
            if mn>con[i+1][0]+dp[2**(v-1)-1][i]:
                mn=con[i+1][0]+dp[2**(v-1)-1][i]



    if mn==inf : print(-1)
    else : print(mn)
    #print(dp)

main()

