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: Time Limit Exceeded)

#include<stdio.h>
#include<stdlib.h>
int *a, *MAXGROUP;
int findRoot(int x){
  if (a[x]<0) return x;
  else return findRoot(a[x]);
}
void unite( int c, int b, int *p ){
  int q=findRoot(c);
  int w=findRoot(b);
  if (w!=q){
    a[q]+=a[w];
    a[w]=q;
    if (*p>a[q]) *p=a[q];
  }
}
int main(){
  int m,n,k1,k2,max=-1;
  MAXGROUP=&max;
  scanf("%d%d",&n,&m);
  int i;
  a=(int*)malloc(n*sizeof(int));
  for (i=0;i<n;i++) a[i]=-1;
  for (i=0;i<m;i++){
    scanf("%d%d",&k1,&k2);
    k1--;k2--;
    unite(k1,k2,MAXGROUP);
  }
  free(a);
  printf("%d\n",-max);
}