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: Time Limit Exceeded)

import sys
from collections import defaultdict

graph = defaultdict(list)

n, m = map(int, sys.stdin.readline().split())

if m == 0:
    print(1)
    exit()

for _ in range(m):
    a,b = map(int, sys.stdin.readline().split())

    if not b in graph[a]:
        graph[a].append(b)
    if not a in graph[b]:
        graph[b].append(a)

visited = set()
maxi = 0

def dfs(start):
    stk = [start]
    visited.add(start)
    temp = 1

    while stk:
        cur = stk.pop()

        for c in graph[cur]:
            if not c in visited:
                visited.add(c)
                stk.append(c)
                temp += 1

    return temp

for j in graph:
    if not j in visited:
        cnt = dfs(j)
        maxi = max(cnt, maxi)

print(maxi)