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: Time Limit Exceeded)

#include <iostream>
#include <stdio.h>
#include <vector>
#include <set>
#include <map>
#include <string.h>
using namespace std;

int dfs(int i, vector<bool> &visited, std::map<int, std::set<int> > friends){
  int result = 1;
  for(std::set<int>::iterator it = friends[i].begin(); it != friends[i].end(); ++it){
    if(!visited[*it]){
      visited[*it] = true;
      result += dfs(*it, visited, friends);
    }
  }
  return result;
}

int main()
{
  int n = 0;
  int m = 0;
  int a = 0;
  int b = 0;
  scanf("%d %d", &n, &m);
  vector<bool> visited(n, false);
  std::map<int, std::set<int> > friends;
  for(int i=0; i<m; i++){
    scanf("%d %d", &a, &b);
    friends[a-1].insert(b-1);
    friends[b-1].insert(a-1);
  }
  int max_size = 1;
  for(int i=0; i<n; i++){
    if(!visited[i]){
      visited[i] = true;
      int size = dfs(i, visited, friends);
      max_size = std::max(size, max_size);
//      printf("%d\n", i);
    }
  }
  printf("%d\n", max_size);
  
//  for(int i=0; i<n; i++){
//    for(int j=0; j<k+1; j++){
//      printf("%lld ", arr[i][j]);
//    }
//    printf("\n");
//  }
//  printf("%lld\n", (arr[n-1][k] - arr[n-1][k-1] + base) % base);
}