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 main(){
   int i,j,ans,n,m,g[200000],c[200000],a,b,cp;

   scanf("%d %d",&n,&m);
   for (i=0;i<n;i++) g[i]=-1;
   cp=0;
   for (i=0;i<m;i++) {
      scanf("%d %d",&a,&b);
      a--;b--;
      if (g[a]<0 && g[b]<0) {
         g[a]=g[b]=cp;
         c[cp]=2;
         cp++;
      } else if (g[a]<0 && g[b]>=0) {
         g[a]=g[b];
         c[g[b]]++;
      } else if (g[a]>=0 && g[b]<0) {
         g[b]=g[a];
         c[g[a]]++;
      } else if (g[a]!=g[b]) 
         if (c[g[a]]<c[g[b]]) {
            for (j=0;j<n;j++)
               if (g[j]==g[a]) g[j]=g[b];
            c[g[b]]+=c[g[a]];
            c[g[a]]=0;
         } else {
            for (j=0;j<n;j++)
               if (g[j]==g[b]) g[j]=g[a];
            c[g[a]]+=c[g[b]];
            c[g[b]]=0;
         }
   }
   ans=c[0];
   for (i=1;i<cp;i++)
      if (ans<c[i]) ans=c[i];
   printf("%d\n",ans);
   return 0;
}