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 <stdlib.h>

int t[300000];
int c[300000];

int root(int x){
  if(t[x]==x) return x;
  return t[x]=root(t[x]);
}

int main(){
  int N,M;
  scanf("%d %d",&N,&M);
  for(int i=1;i<=N;i++){
    t[i]=i;
    c[i]=1;
  }
  int A,B,RA,RB;
  for(int i=1;i<=M;i++){
    scanf("%d %d",&A,&B);
    if(A>B){
      A^=B;
      B^=A;
      A^=B;
    }
    RA=root(A);
    RB=root(B);
    if(RA!=RB){
      t[B]=A;
      c[A]+=c[B];
    }
  }
  int Ans=0;
  for(int i=1;i<=N;i++){
    if(root(i)==i)
      if(Ans<c[i])
        Ans=c[i];
  }
  printf("%d\n",Ans);
  return 0;
}