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 <bits/stdc++.h>
using namespace std;
#define pi 2*asin(1.0)
#define MAX 10000000
#define MOD 1000000007


 
//Answer
 
 
int main() {

long N,M,t;
typedef pair<long,long> P;
cin>>N>>M;
vector<P> ab(M);
vector<P> ba(M);
vector<long> X(N,1);
for(long i=0;i<M;i++){
cin>>ab.at(i).first;
cin>>ab.at(i).second;
if(ab.at(i).first>ab.at(i).second){
    t=ab.at(i).first;
    ab.at(i).first=ab.at(i).second;
    ab.at(i).second=t;
}
	ba.at(i).first=ab.at(i).second;
    ba.at(i).second=ab.at(i).first;
}
sort(ab.begin(),ab.end());
sort(ba.begin(),ba.end());
queue<long> que;
long count,max=0;
for(long i=0;i<N;i++){
count=0;
que.push(i+1);
while(que.size()){
    long q=que.front();
    que.pop();
  cout<<q<<endl;
    if(X.at(q-1)==1){
        count++;
        X.at(q-1)=0;
      P pp=make_pair(q,0);
       auto itr1=lower_bound(ab.begin(),ab.end(),pp);
       auto itr2=lower_bound(ba.begin(),ba.end(),pp);
      long it=distance(ab.begin(),itr1);
      long it2=distance(ba.begin(),itr2);
   //   cout<<distance(ab.begin(),itr1)<<endl;
      while(it<M&&(q==ab.at(it).first)){
         que.push(ab.at(it).second);
            it++;
       }
      while(it2<M&&(q==ba.at(it2).first)){
         que.push(ba.at(it2).second);
            it2++;
       }
      
    }
}
if(max<count){
    max=count;
}
}
cout<<max<<endl;

}