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>

int up(const void *a,const void *b){
  return *(int *)a - *(int *)b;
}

int main(void){
  int n , m;
  scanf("%d %d",&n ,&m);

  int a[m] , b[m];
  int pass[400000][2];

  for(int i = 0;i < m;i ++){
    scanf("%d %d",&a[i],&b[i]);

    if(a[i] > b[i]){
      int temp = a[i];
      a[i] = b[i];
      b[i] = temp;
    }
    pass[i][0] = a[i];
    pass[i][1] = b[i];
  }

 
  int count[200001] = {0};
  int x = 0 , y = 0;
  for(int i = 0;i < m;i ++){
    if(pass[i][0] == x && pass[i][1] == y);
      else{
        count[pass[i][0]]++;
      count[pass[i][1]]++;
      x = pass[i][0];
      y = pass[i][1];
      }
  }

  int max = 0;
  for(int i = 0;i < n;i ++){
    if(max < count[i]){
      max = count[i];
    }
  }

  printf("%d\n",max+1);

  return 0;

}