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: Time Limit Exceeded)

n, m = map(int, input().split())
a, b = [], []
for i in range(m):
  _, __ = map(int, input().split())
  _ -= 1
  __ -= 1
  a.append(_)
  b.append(__)
  
# union-find

# 誰の肩に手を置いているか（先頭の場合は-1）
parent = [-1 for i in range(n)]

# 列車の深さ
depth = [0 for i in range(n)]

# kさんの「運転手」が誰かを調べる関数
def find_root(k):
  while parent[k] != -1:
    k = parent[k]
  return k 

for i in range(m):
  # a[i] と b[i] の合併
  root_a = find_root(a[i])
  root_b = find_root(b[i])
  
  # 浅い列車を持っている方が負けるようにする
  if depth[root_a] > depth[root_b]:
    root_a, root_b = root_b, root_a
  parent[root_a] = root_b
  depth[root_b] = max(depth[root_a]+1, depth[root_b])
  
group = [find_root(i) for i in range(n)]

from collections import Counter
print(Counter(group).most_common()[0][1])