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)

from collections import Counter

class UnionFind():
    def __init__(self, n):
        self.par = [i for i in range(n)]
        self.rank = [0] * n
        
    def find(self, x):
        if self.par[x] == x:
            return x
        else:
            self.par[x] = self.find(self.par[x])
            return self.par[x]
            
    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)
        
        if self.rank[x] < self.rank[y]:
            self.par[x] = y
        else:
            self.par[y] = x
            
            if self.rank[x] == self.rank[y]:
                self.rank[x] += 1
            

N, M = map(int, input().split())
UF = UnionFind(N)
for i in range(M):
    a, b = map(int, input().split())
    a -= 1
    b -= 1
    big = max(a, b)
    small = min(a, b)
    
    if UF.par[big] == big:
        UF.par[big] = small
    else:
        UF.union(big, small)
        
for i in range(N):
    _ = UF.find(i)

c = Counter(UF.par)
print(max(c.values()))