Problem Statement

There is a simple undirected graph with N vertices and M edges. The edges are initially painted white.
The vertices are numbered 1 through N, and the edges are numbered 1 through M.
Edge i connects vertex A_i and vertex B_i, and the cost required to paint it black is C_i.
"Making a Q" means painting four or more edges so that:

• all but one of the edges painted black form a simple cycle, and
• the edge painted black not forming the cycle connects a vertex on the cycle and another not on the cycle.

Determine if one can make a Q. If one can, find the minimum total cost required to make a Q.

Input

The input is given from Standard Input in the following format:
N M
A_1 B_1 C_1
A_2 B_2 C_2
$\vdots$
A_M B_M C_M

Output

If one can make a Q, print the minimum total cost required to do so; otherwise, print -1.

Constraints

• $4\leq N \leq 300$
• $4\leq M \leq \frac{N(N-1)}{2}$
• $1 \leq A_i < B_i \leq N$
• $(A_i,B_i) \neq (A_j,B_j)$, if $i \neq j$.
• $1 \leq C_i \leq 10^5$
• All input values are integers.

Sample Input 1

5 6
1 2 6
2 3 4
1 3 5
2 4 3
4 5 2
3 5 1

Sample Output 1

15

By painting edges 2,3,4,5, and 6,

• edges 2,4,5, and 6 forms a simple cycle, and
• edge 3 connects vertex 3 (on the cycle) and vertex 1 (not on the cycle),

so one can make a Q with a total cost of 4+5+3+2+1=15.
Making a Q in another way costs 15 or greater, so the answer is 15.

Sample Input 2

4 4
1 2 1
2 3 1
3 4 1
1 4 1

Sample Output 2

-1

Sample Input 3

6 15
2 6 48772
2 4 36426
1 6 94325
3 6 3497
2 3 60522
4 5 63982
4 6 4784
1 2 14575
5 6 68417
1 5 7775
3 4 33447
3 5 90629
1 4 47202
1 3 90081
2 5 79445

Sample Output 3

78154

见题面 Input 部分

见题面 Output 部分