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<stdio.h>
#include<string.h>
int main(){
 	int b,c,i,j,n,m;	
  	int ans = 0;
  	int a[220001] = {0};
  int d[220001]={0};
 
  scanf("%d %d",&n,&m);
  for(i=0;i<n;i++){
   a[i] = i; 
  }
  for(i=0;i<m;i++){
   scanf("%d %d",&b,&c);
    b--;
    c--;
    if(b<c){
        a[c]=a[b];
    }else{
       a[b]=a[c]; 
    }
    
    
  }
  for(i=0;i<n;i++){
   d[a[i]]++;
  }
  for(i=0;i<n;i++){
   if(ans<d[i]){
    ans=d[i]; 
   }
  }
  
  printf("%d\n",ans);
  return 0;
}