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: Time Limit Exceeded)

import collections as col
from collections import defaultdict as dd

d=dd(list)

N,M=map(int, input().split())
p=[-1]*N
cnt=[0]*N
p[0]=1
cnt[0]=1
for _ in range(M):
    i,j=map(int, input().split())
    if i>j:
        i,j=j,i
    d[i].append(j)

for i in d:
    for j in d[i]:
        if p[i-1]==-1:
            if p[j-1]==-1:
                if i<j:
                    p[i-1]=i
                    p[j-1]=i
                else:
                    p[i-1]=j
                    p[j-1]=j
            else:
                p[i-1]=p[j-1]
        else:
            if p[j-1]==-1:
                p[j-1]=p[i-1]
            else:
                if p[i-1]<p[j-1]:
                    temp1=p[j-1]
                    temp2=p[i-1]
                else:
                    temp1=p[j-1]
                    temp2=p[i-1]
                p=[temp2 if k==temp1 else k for k in p]

ans=col.Counter(p)

if ans.most_common()[0][0]==-1:
    print(ans.most_common()[1][1])
else:
    print(ans.most_common()[0][1])