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 PyPy3 (7.3.0) (judge verdict: Wrong Answer)

import math
import queue

n, m = map(int, input().split(" "))
abL = [list(map(int, input().split(" "))) for _ in range(m)]
visited = [False for _ in range(n + 1)]

d = {}

for a, b in abL:
    if a in d:
        d[a].add(b)
    else:
        d[a] = set([b])

    if b in d:
        d[b].add(a)
    else:
        d[b] = set([a])

ans = 0
for k, v in d.items():
    if not visited[k]:
        cnt = 0
        q = queue.Queue()
        q.put(k)
        while q.qsize() != 0:
            ss = q.get()
            cnt += 1
            if cnt >= n:
                break
            visited[ss] = True
            for dd in list(d[ss]):
                if not visited[dd]:
                    q.put(dd)
        ans = max(ans, cnt)

print(ans)