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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#include<iostream>
using namespace std;
int a[200005];
int cnt[200005];
int findroot(int v){
	return a[v]==v?v:a[v]=findroot(a[v]);
}
int main(){
	int n,m;
	cin>>n;
	cin>>m;
	for(int i=1;i<=n;i++){
		a[i]=i;
		cnt[i]=1;
	}
	int max=0;
	for(int i=0;i<m;i++){
		int x,y;
		cin>>x>>y;
		int findrootx=findroot(x);
		int findrooty=findroot(y);
		if(findrootx!=findrooty){
			a[findrooty]=findrootx;
			cnt[findrootx]+=cnt[findrooty];
		}
	}
	for(int i=1;i<=n;i++){
		if(findroot(i)==i){
			if(max<cnt[i]){
				max=cnt[i];
			}
		}
	}
	cout<<max;
	
}