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: Accepted)

#include <stdio.h>

int r[200000];

int root(int i) {
   if (r[i]<0) return i;
   else        return root(r[i]);
}

int main(){
   int i,ans,n,m,a,b,ra,rb;

   scanf("%d %d",&n,&m);
   for (i=0;i<n;i++) r[i]=-1;
   for (i=0;i<m;i++) {
      scanf("%d %d",&a,&b);
      a--;b--;
      ra=root(a);rb=root(b);
      if (ra!=rb) 
         if (r[ra]>r[rb]) {
            r[rb]+=r[ra];
            r[ra]=rb;
         } else {
            r[ra]+=r[rb];
            r[rb]=ra;
         }
   }
   ans=0;
   for (i=0;i<n;i++)
      if (ans>r[i]) ans=r[i];
   printf("%d\n",-ans);
   return 0;
}