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: Runtime Error)

#include<stdio.h>
int p[100010];
int d[100010]={};
int S[100010];
int Top(int a){
  for(;p[a]-a;a=p[a]);
  return a;
}
int Find(int a,int b){
  int n,r;
  for(r=Top(a);p[a]!=a;a=n){
    n=p[a];
    p[a]=r;
  }
  for(r=Top(b);p[b]!=b;b=n){
    n=p[b];
    p[b]=r;
  }
  return a-b?1:0;
}
void Union(int a,int b){
  if(Find(a,b)){
    a=Top(a);
    b=Top(b);
    d[a]<d[b]?(p[a]=b):(p[b]=a);
    if(d[a]==d[b])d[a]++;
    S[a]=S[b]=S[a]+S[b];
  }
}
int main(){
  int i,n,m,a,b;
  scanf("%d %d",&n,&m);
  for(i=0;i<=n;i++)p[i]=i;
  for(i=0;i<=n;i++)S[i]=1;
  while(m--){
    scanf("%d %d",&a,&b);
    Union(a,b);
  }
  for(i=0;i<n;i++){
    if(m<S[i+1])m=S[i+1];
  }
  printf("%d\n",m);
  return 0;
}