Problem:
Score : 400 points

Problem Statement

There are N persons called Person 1 through Person N.

You are given M facts that "Person A_i and Person B_i are friends." The same fact may be given multiple times.

If X and Y are friends, and Y and Z are friends, then X and Z are also friends. There is no friendship that cannot be derived from the M given facts.

Takahashi the evil wants to divide the N persons into some number of groups so that every person has no friend in his/her group.

At least how many groups does he need to make?

Constraints

2 \leq N \leq 2\times 10^5

0 \leq M \leq 2\times 10^5

1\leq A_i,B_i\leq N

A_i \neq B_i

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
\vdots
A_M B_M

Output

Print the answer.

Sample Input 1

5 3
1 2
3 4
5 1

Sample Output 1

3

Dividing them into three groups such as \{1,3\}, \{2,4\}, and \{5\} achieves the goal.

Sample Input 2

4 10
1 2
2 1
1 2
2 1
1 2
1 3
1 4
2 3
2 4
3 4

Sample Output 2

4

Sample Input 3

10 4
3 1
4 1
5 9
2 6

Sample Output 3

3

Solution in Python (3.8.2) (judge verdict: Wrong Answer)

import sys
input = sys.stdin.readline

n,m=map(int,input().split())
p = [tuple(sorted(list(map(int,input().split())))) for ii in range(m)]
p=set(p)

m=len(p)

g=[-1]*(n+1)
r=0

p2=list()

rr=[2]

import numpy as np
for pii in p:
  if g[pii[0]]<0 and g[pii[1]]<0:
    g[pii[0]]=r
    g[pii[1]]=r
    if r>0:
      rr.append(2)
    r+=1
  elif g[pii[0]]>=0 and g[pii[1]]>=0:
    g0=g[pii[0]]
    g1=g[pii[1]]
    if g0!=g1:
      p2.append((g[pii[0]],g[pii[1]]))
  elif g[pii[0]]>=0:
    g[pii[1]]=g[pii[0]]
    rr[g[pii[0]]]+=1
  elif g[pii[1]]>=0:
    g[pii[0]]=g[pii[1]]
    rr[g[pii[1]]]+=1

for p2ii in p2:
  g0=g[p2ii[0]]
  g1=g[p2ii[1]]
  rr[g1]+=rr[g0]
  rr[g0]=0

#print(rr)
print(max(rr))