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: Accepted)

import sys

sys.setrecursionlimit(10 ** 7)
rl = sys.stdin.readline


class UnionFind:
    def __init__(self, n: int):
        self.n = n
        self.parents = [-1] * n
    
    def find(self, x: int):
        if self.parents[x] < 0:
            return x
        else:
            self.parents[x] = self.find(self.parents[x])
            return self.parents[x]
    
    def union(self, x: int, y: int):
        x = self.find(x)
        y = self.find(y)
        
        if x == y:
            return
        if self.parents[y] < self.parents[x]:
            x, y = y, x
        
        self.parents[x] += self.parents[y]
        self.parents[y] = x
    
    def size(self, x: int):
        return -self.parents[self.find(x)]
    
    def same(self, x: int, y: int):
        return self.find(x) == self.find(y)
    
    def members(self, x: int):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]
    
    def roots(self):
        return [i for i, x in enumerate(self.parents) if x < 0]
    
    def group_count(self):
        return len(self.roots())
    
    def all_group_members(self):
        return {r: self.members(r) for r in self.roots()}
    
    def __str__(self):
        return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())


def solve():
    N, M = map(int, rl().split())
    uf = UnionFind(N)
    for _ in range(M):
        a, b = map(lambda n: int(n) - 1, rl().split())
        uf.union(a, b)
    
    ans = 0
    for i in range(N):
        ans = max(ans, uf.size(i))
    print(ans)


if __name__ == '__main__':
    solve()