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)

/*
	Name:
	Author: xiaruize
	Date:
*/
#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
#define ull unsigned long long
#define MOD 1000000007
#define ALL(a) (a).begin(), (a).end()
#define forn(i, n) for (int i = 0; i < int(n); i++)
#define form(i,j,n) for(int i=int(j);i<=int(n);i++)
#define pb push_back
#define mk make_pair
#define pii pair<int,int>
#define pis pair<int,string>
#define sec second
#define ll long long
#define fir first
#define sz(a) int((a).size())

int fa[200005];
int cntf[200005];
int ans=0;
int get(int x)
{
	if(x==fa[x])
		return x;
	return fa[x]=get(fa[x]);
}

void _merge(int a,int b)
{
	int x=get(a),y=get(b);
	if(x==y)
		return ;
	fa[x]=y;
	cntf[y]+=cntf[x];
	cntf[x]=0;
	ans=max(ans,cntf[y]);
}

signed main()
{
 	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
 	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	int n,m;
	cin>>n>>m;
	for(int i=0;i<=n+1;i++)
	{
		fa[i]=i;
		cntf[i]=1;
	}
	int a,b;
	for(int i=0;i<m;i++)
	{
		cin>>a>>b;
		_merge(a,b);
	}
	if(ans==0)
		ans=1;
	cout<<ans<<endl;
	return 0;
}