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)

/*-- ILSH  --*/
#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
typedef long double ld;
#define pb push_back 
#define pop pop_back
#define mp make_pair
#define vii vector < ll >
#define vll vector < ll > 
#define dt cout<<"HERE\n";
#define pii pair < ll , ll >
#define pll pair < ll , ll >
#define vpi vector < pii >
#define vpl vector < pll >
#define fi first
#define se second 
#define sz size 
#define len length 
const ll inf=1e18+1e17;
const ll moda =29;
const ll modb =34483;
//const ll mod = 998244353;

ll modInverse(ll a,ll m){ll m0=m;ll y=0,x=1;if(m == 1)return 0;while(a> 1){ll q=a/m;ll t=m;m=a%m,a=t;t=y;y=x-q*y;x=t;}if(x<0)x+=m0;return x;}  
/*ll powm(ll a,ll b){a=a%mod;ll res=1;while(b){if(b&1)res=(res*a)%mod;a=(a*a)%mod;b>>=1;}return (res%mod);}*/
const ll N =2e5+5;
int arr[N],brr[N];
void init(){
	for ( int i=0;i<N;i++){
		arr[i]=i;
		brr[i]=1;
	}
}
int root( int pos){
	if ( arr[pos]==pos)
		return pos;
	return arr[pos]=root( arr[pos]);
}
void join ( int x, int y){
	x=root( x);
	y=root( y);
	if ( x==y)
		return ;
	if ( brr[x]> brr[y])
		swap( x,y);
	arr[x]=y;
	brr[y]+=brr[x];
}
void solve(){
	int n,m;
	cin>>n>>m;
	init();
	for ( int i=0;i<m;i++){
		int aa, bb;
		cin>>aa>>bb;
		aa--;bb--;
		join ( aa,bb);
	}
	int mx=0;
	for ( int i=0;i<n;i++)
		mx=max(brr[root(i)],mx);
	cout<<mx<<'\n';

}
int main(){
	ios_base::sync_with_stdio(false);
	cin.tie( NULL);
	cout.tie(NULL);

	#ifndef ONLINE_JUDGE
		freopen ("INPUT.txt" , "r" , stdin);
		freopen ("OUTPUT.txt", "w" , stdout);
	#endif
	
	ll t=1;
	//cin>>t;
	for ( ll i=1;i<=t;i++){
		//cout<<"Case #"<<i<<": ";
    	solve();
    }
}