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<bits/stdc++.h>   
#include<tr1/unordered_map>
#define ll long long int  
#define mkp make_pair 
#define mod 1000000007 
#define maxn 100001 
#define ff first
#define ss second 
using namespace std;    
int par[2*maxn]={0},s[2*maxn]={0}; 
int find(int a){  
    if(par[a]==a) 
       return a; 
    return par[a]=find(par[a]);
}  
void merge(int a,int b){ 
      int p=find(a); 
      int q=find(b); 
      if(p!=q){  
         if(s[p]>=s[q]){  
               par[q]=p; 
               s[p]+=s[q];  
               s[q]=0;
          } 
         else{  
            par[p]=q; 
            s[q]+=s[p]; 
            s[p]=0;
           }
      }
    }
int main(){    
  ios_base::sync_with_stdio(false); 
  cin.tie(NULL);    
  int n,m,a,b; 
  cin>>n>>m; 
  for(int i=0;i<=n;++i)  
     par[i]=i,s[i]=1;  

  while(m--){  
      cin>>a>>b; 
      merge(a,b);
  }  
  int ans=0;
  for(int i=1;i<=maxn;++i){   
     ans=max(ans,s[i]);
        
    }   
  cout<<ans<<"\n";
  
  
  return 0;
}