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: Wrong Answer)

#include<iostream>
#include<cstdio>
using namespace std;
int n,m,f[300006],a,b,ans=0,max1[300006];
int find(int x)
{
	return x==f[x]?x:(f[x]=find(f[x]));
}
void merge(int i,int j)
{
	f[find(i)]=find(j);
	max1[f[j]]+=max1[f[i]]+1;
	if(ans<max1[f[j]])
	ans=max1[f[j]];
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;++i)
	{
		f[i]=i;
	}
	for(int i=1;i<=m;++i)
	{
		scanf("%d%d",&a,&b);
		if(find(a)!=find(b))
		merge(a,b);
	}
	ans++;
	printf("%d",ans);
	return 0;
}