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 <iostream>
#include <vector>
#include <set>
#include <algorithm>
using namespace std;
#define rep(i, n)for(int i=0; i<n; i++)

int main(){
  int N, M;
  cin >> N >> M;
  vector<set<int> > G(N);
  int a, b;
  rep(i, M){
    cin >> a >> b;
    G[a-1].insert(b-1);
    G[b-1].insert(a-1);
  }

  int max = 0;
  int tmp;
  rep(i, N){
    tmp = (G[i].size()+1);
    
    if(max < tmp)
      max = tmp ;
  }

  /*rep(i, N){
    cout<< i+1<<" ";
    for(auto p : G[i])
      cout << p+1 << ", ";
    cout<<endl;
    }*/
  
  cout << max << endl;
  
  return 0;
}