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)

/*author @dhanush*/

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define boost ios_base::sync_with_stdio(false);cin.tie(NULL)
#define pb push_back
#define se second
#define fi first
#define INF INT_MAX

int parent[200001];

int find(int a)
{
  if(parent[a]<0)
  {
    return a;
  }
  else
  {
    int x=find(parent[a]);
    parent[a]=x;
    return x;
  }

}

void Union(int a , int b)
{
  parent[a] += parent[b];   //adding size of set b to set a
  parent[b] = a;        //making a , parent of new set
}


int main()
{


  int n , m , a , b;
  cin >> n >> m;

  for (int i = 1; i <= n; i++)
    parent[i] = -1;   //initializing

  while (m--)
  {
    cin >> a >> b;
    a = find(a) , b = find(b);

    if (a != b) Union(a , b);
  }

  int res = -1;

  for (int i = 1; i <= n; i++)
    if (parent[i] < 0)
      res = min(res, parent[i]);

  cout << abs(res);



}