10
16
2015
0

[Codeforces Round #318] Bear and Three Musketeers


B. Bear and Three Musketeers
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Do you know a story about the three musketeers? Anyway, you will learn about its origins now.

Richelimakieu is a cardinal in the city of Bearis. He is tired of dealing with crime by himself. He needs three brave warriors to help him to fight against bad guys.

There are n warriors. Richelimakieu wants to choose three of them to become musketeers but it's not that easy. The most important condition is that musketeers must know each other to cooperate efficiently. And they shouldn't be too well known because they could be betrayed by old friends. For each musketeer his recognition is the number of warriors he knows, excluding other two musketeers.

Help Richelimakieu! Find if it is possible to choose three musketeers knowing each other, and what is minimum possible sum of their recognitions.

Input

The first line contains two space-separated integers, n and m (3 ≤ n ≤ 40000 ≤ m ≤ 4000) — respectively number of warriors and number of pairs of warriors knowing each other.

i-th of the following m lines contains two space-separated integers ai and bi (1 ≤ ai, bi ≤ nai ≠ bi). Warriors ai and bi know each other. Each pair of warriors will be listed at most once.

Output

If Richelimakieu can choose three musketeers, print the minimum possible sum of their recognitions. Otherwise, print "-1" (without the quotes).

Sample test(s)
input
5 6
1 2
1 3
2 3
2 4
3 4
4 5
output
2
input
7 4
2 1
3 6
5 1
1 7
output
-1
Note

In the first sample Richelimakieu should choose a triple 123. The first musketeer doesn't know anyone except other two musketeers so his recognition is 0. The second musketeer has recognition 1 because he knows warrior number 4. The third musketeer also has recognition 1 because he knows warrior 4. Sum of recognitions is 0 + 1 + 1 = 2.

The other possible triple is 2, 3, 4 but it has greater sum of recognitions, equal to 1 + 1 + 1 = 3.

In the second sample there is no triple of warriors knowing each other.

题目大意:有n个人每个人,他们有些人互相认识。从他们中选出互相认识的3个人,定义他们中一个人的“认识程度”为他认识的人数(不包括另外两个人),要求这3个人的“认识程度”之和最小。输出这个最小的“认识程度”。


枚举3个人要超时,变成枚举一条边和另外一个人就行了。

代码在此。

Category: 题解 | Tags: Codeforces 暴力 | Read Count: 489

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