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 <bits/stdc++.h>
using namespace std;

class Union_Find{
private:
  vector<int64_t> par;
  vector<int64_t> rank;
  vector<int64_t> num;

  int64_t find(int64_t x){
    if(par.at(x)==x){
      return x;
    }else{
      return par.at(x)=find(par.at(x));
    }
  }

public:
  Union_Find(int64_t n){
    par.resize(n);
    rank.resize(n,0);
    num.resize(n,1);
    for(int64_t i=0;i<n;i++){
      par.at(i)=i;
    }
  }

  void unite(int64_t x,int64_t y){
    x=find(x);
    y=find(y);
    if(x==y){
      return;
    }
    if(rank.at(x)<rank.at(y)){
      par.at(x)=y;
      num.at(y)+=num.at(x);
    }else if(rank.at(x)>rank.at(y)){
      par.at(y)=x;
      num.at(x)+=num.at(y);
    }else{
      par.at(x)=y;
      rank.at(y)++;
      num.at(y)+=num.at(x);
    }
  }

  bool same(int64_t x,int64_t y){
    return find(x)==find(y);
  }

  int64_t size(int64_t x){
    return num.at(find(x));
  }

};

int main(){
  int N,M;
  cin >> N >> M;
  int A,B;
  Union_Find uf(N+1);
  for(int i=0;i<M;i++){
    cin >> A >> B;
    uf.unite(A,B);
  }
  int64_t ans=0;
  for(int i=1;i<=N;i++){
    ans=max(ans,uf.size(i));
  }
  cout << ans << endl;
  return 0;
}