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)

from collections import *
N,M=map(int,input().split())
AB=[list(map(int,input().split())) for i in range(M)]

lst=[[] for i in range(N+1)]

for a,b in AB:
    lst[a].append(b)
    lst[b].append(a)



used=set(list(range(1,N+1)))
value=-10**18
while used:
    a=used.pop()
    stack=[a]

    visited={a}
    while stack:
        a=stack.pop()
        for p in lst[a]:
            if not p in visited:
                visited.add(p)
                stack.append(p)
    value=max(len(visited),value)
print(value)