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: Runtime Error)

import numpy as np

N, M = map(int, input().split())

members = np.array([-1] * N).astype(np.int)

group_count = 0
groups = [0] * N
group_dict = {}
for m in range(M):
    A, B = map(int, input().split())
    A = A - 1
    B = B - 1
    if members[A]==-1 and members[B]==-1:
        members[A] = members[B] = group_count
        groups[group_count] = 2
        group_dict[group_count] = [A, B]
        group_count += 1
    
    elif members[A]!=-1 and members[B]==-1:
        members[B] = members[A]
        groups[members[A]] += 1
        group_dict[members[A]].append(B)
    
    elif members[B]!=-1 and members[A]==-1:
        members[A] = members[B]
        groups[members[B]] += 1
        group_dict[members[B]].append(A)
    
    else:
        if members[A] != members[B]:
            old = members[B]
            groups[members[A]] += groups[members[B]]
            groups[members[B]] = 0
            group_dict[members[A]] = group_dict[members[A]] +  group_dict[members[B]]
            for aaaaaa in group_dict[members[B]]:
                members[aaaaaa] = members[A]
            group_dict[old] = []

ans = max(groups)
if ans == 0:
    print(1)
else:
    print(ans)