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;
#define ll long long int

int parent(int a[],int x){
    if(a[x]==x) return x;
    else return a[x]=parent(a,a[x]);
}

void unite(int x,int y,int a[]){
    int dx=parent(a,x);
    int dy=parent(a,y);
    if(dx!=dy)
    a[dx]=dy;
}

int main(){
   ll n,m;
   cin>>n>>m;
   int a[n+1];
   for(int i=1;i<n+1;i++)
   a[i]=i;
   while(m--){
     ll u,v;
     cin>>u>>v;
     unite(u,v,a);
   }
   unordered_map<int,int>mp;
   for(int i=1;i<=n;i++){
       //cout<<parent(a,i)<<" ";
       mp[parent(a,i)]++;
   }
   int maxi=INT_MIN;
   for(auto i:mp)
    maxi=max(maxi,i.second);
    cout<<maxi;
   return 0;
}