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<iostream>
#include<vector>
#include<string>
#include <queue>
#include<map>
#include<set>
#include <cstring>
using namespace std;
#define ll long long int

#define mod 1000000007

vector<vector<ll>> g(200001) ;
ll cnt = 0 ;

void dfs(ll u , vector<bool> &vis)
{
    vis[u] = true ;
    cnt++ ;

    for(ll i = 0 ; i < g[u].size() ; i++)
    {
        if(!vis[g[u][i]])
            dfs(g[u][i] , vis) ;
    }
}
int main()
{
   ll n , m ;
   cin>>n>>m ;

   for(ll i = 0 ; i < m ; i++)
   {
       ll u , v ;
       cin>>u>>v ;

       g[u].push_back(v) ;
       g[v].push_back(u) ;
   }
   vector<bool> vis(n + 1 , false) ;
   ll ans = 0 ;
   for(ll i = 1 ; i <= n ; i++)
   {
       if(!vis[i])
       {
           dfs(i , vis) ;
           ans = max(ans , cnt) ;
           cnt = 0 ;
       }
   }

   cout<<ans ;

}