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: Accepted)

# -*- coding: utf-8 -*-
"""
Created on Thu Sep  3 20:46:45 2020

@author: liang
"""


class UnionFind():
    def __init__(self, N):
        self.N = N
        self.par = [-1]*N
    
    def find(self, x):
        if self.par[x] < 0:
            return x
        self.par[x] = self.find(self.par[x])
        return self.par[x]
    
    def union(self, a, b):
        a = self.find(a)
        b = self.find(b)
        if a == b:
            return self
        if self.par[a] > self.par[b]:
            a, b = b, a
        self.par[a] += self.par[b]
        self.par[b] = a
    
    def same(self, a, b):
        a = self.find(a)
        b = self.find(b)
        return a == b
    
N, M = map(int, input().split())
uf = UnionFind(N)
for i in range(M):
    a, b = map(int,input().split())
    uf.union(a-1, b-1)

ans = -min(uf.par)
print(ans)