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;
#define ll long long int
#define pb push_back
#define mk make_pair
#define mod 1000000007
int find(vector<int> pare,int i){
    if(pare[i]==i)
     return i;
    return find(pare,pare[i]);
}
void uni(vector<int> &pare,int i,int j){
    int xset=find(pare,i);
    int yset=find(pare,j);
    if(xset!=yset)
    pare[xset]=yset;
   return;
}
int main(){
  int n,t;
  cin>>n>>t;
  vector<int> pare(n+1);
  for(int i=1;i<=n;i++)
   pare[i]=i;
  for(int i=0;i<t;i++){
      int x,y;
      cin>>x>>y;
      uni(pare,x,y);
  }
  vector<int> hash(1000001,0);
  int ans=0;
  for(int i=1;i<=n;i++){
      if(pare[i]!=-1){
       hash[pare[i]]++;
       ans=max(ans,hash[pare[i]]);}
  }
 // sort(hash.rbegin(),hash.rend());
  cout<<ans+1<<endl;
}