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;

int main(){
  int n,m;
  cin>>n>>m;
  vector<vector<int>> info(m,vector<int>(2));
  vector<int> index(m);
  for(int i=0;i<m;i++){
    cin>>info.at(i).at(0)>>info.at(i).at(1);
    index.at(i)=i;
  }
  set<int> friends;
  bool next=false;
  int max=1;
  while(index.size()>0){
    friends={info.at(index.at(0)).at(0),info.at(index.at(0)).at(1)};
    index.erase(index.begin());
    while(!next){
      next=true;
      /*
      for(int i=index.size()-1;i>=0;i--){
        if(friends.count(info.at(index.at(i)).at(0))){
          friends.insert(info.at(index.at(i)).at(1));
          index.erase(index.begin()+i);
          next=false;
        }else if(friends.count(info.at(index.at(i)).at(1))){
          friends.insert(info.at(index.at(i)).at(0));
          index.erase(index.begin()+i);
          next=false;
        }
      }
      */
      for(int i=0;i<index.size();i++){
        if(friends.count(info.at(index.at(i)).at(0))){
          friends.insert(info.at(index.at(i)).at(1));
          index.erase(index.begin()+i);
          i--;
          next=false;
        }else if(friends.count(info.at(index.at(i)).at(1))){
          friends.insert(info.at(index.at(i)).at(0));
          index.erase(index.begin()+i);
          i--;
          next=false;
        }
      }
    }
    if(friends.size()>max){
      max=friends.size();
    }
  }
  cout<<max<<endl;
}