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: Runtime Error)

import collections
def solve():
  N,M = map(int,input().split())
  ABlist = [[0 for _ in range(2)] for _ in range(M)]
  WA = [0]*N
  for i in range(N):
    WA[i] = i+1
  for i in range(M):
    A, B = map(int,input().split())
    if A>B:
      A,B =B,A
    ABlist[i][0] = A
    ABlist[i][1] = B
  ABlist.sort(key=lambda x: (x[0],x[1]))
  print(ABlist)
  for i in range(M):
    A = ABlist[i][0]
    B = ABlist[i][1]
    if WA[B-1]<WA[A-1]:
      WA[A-1] = WA[B-1]
    else:
      WA[B-1] = WA[A-1]
    print(i, WA, A,B)
  c = collections.Counter(WA)
  print(c.most_common()[0][1])
solve()