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: Wrong Answer)

#Union-Find木の実装
from collections import Counter
N,M=map(int,input().split())

#初期化
par=[i for i in range(N)]#親の要素(par[x]=xのときxは木の根)
rank=[0]*N#木の深さ

#木の根を求める
def find(x):
  if par[x] == x:
    return x
  else:
    par[x] = find(par[x])
    return par[x]

#xとyの属する集合を併合
def unite(x,y):
  x=find(x)
  y=find(y)
  if x == y:
    return
  if rank[x] < rank[y]:
    par[x] = y
  else:
    par[y] = x
    if rank[x] == rank[y]:
      rank[x]+=1

#xとyが同じ集合に属するか否か
def same(x,y):
  return find(x)==find(y)

for _ in range(M):
  a,b = map(int,input().split())
  a,b = a-1, b-1
  unite(a,b)

c=Counter(par)
c=c.most_common()[0][1]
print(c)