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++ (Clang 10.0.0) (judge verdict: Accepted)

#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
int m , n ,f[200005] , l ,sum, x,y,book[200005];
int df(int u)
{
	 if(f[u] ==u ) return u;
	 else 
	 {
	 	 f[u] = df(f[u]);
	 	 return f[u];
	 }
}
void getf(int t1 ,int t2)
{
	int k1 = df(t1) ;
	int k2 = df(t2) ;
	
    f[k2] = k1 ;

}
int main()
{   
    cin >> n >> m ;
	for(int i = 1 ; i <= n; i++) f[i] = i ;
	for(int i = 1; i <= m; i++)
	{ 
	   cin >> x >> y ;
	   getf(x,y);
	}
	
	for(int i = 1 ; i <= n; i++) df(i);
	
	for(int i = 1 ; i <= n; i++)  book[f[i]]++;
	
	for(int i = 1 ; i <= n; i++)
	sum = max(sum,book[i]);
	
	cout << sum ;
	return 0 ; 
}