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: Runtime Error)

import numpy as np

class UnionFind:
    def __init__(self, sz: int):
        self._par: np.array = np.full(sz + 1, -1)

    def root(self, a: int):
        if self._par[a] < 0:
            return a
        self._par[a] = self.root(self._par[a])
        return self._par[a]
    
    def size(self, a: int):
        return -self._par[self.root(a)]
    
    def unite(self, a, b):
        a = self.root(a)
        b = self.root(b)
        if a != b:
            if self.size(a) < self.size(b):
                a, b = b, a
            self._par[a] += self._par[b]
            self._par[b] = a
    
if __name__ == '__main__':
    N, M = map(int, input().split())
    uf = UnionFind(N + 1)
    for i in range(M):
        a, b = map(int, input().split())
        uf.unite(a, b)
    ans = 1
    for i in range(1, N + 1):
        ans = max(ans, uf.size(i))
    print(ans)