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)

///Bismillahir Rahmanir Rahim
#include<bits/stdc++.h>
using namespace std;
#define                          ll                          long long int
ll gcd(int a,int b)
{
    if(b==0)
        return a;
    return gcd(b,a%b);
}

#define                          in(x)                       scanf("%lld",&x)
#define                          in2(x,y)                    scanf("%lld %lld",&x,&y)
#define                          in3(x,y,z)                  scanf("%lld %lld %lld",&x,&y,&z)
#define                          out(x)                      printf("%lld\n",x)
#define                          cy                          printf("YES\n")
#define                          cn                          printf("NO\n")
#define                          inf                         1e18
#define                          neg                         -1e18
#define                          mx                          200009
#define                          fast                        ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define                          fi                          first
#define                          si                          second
#define                          ce                          cout<<endl
#define                          pb                          push_back
#define                          vc(x)                       x.begin(),x.end()
#define                          pb                          push_back
#define                          pi                          pair<ll,ll>
#define                          debug                       printf("ok\n");
#define                          pg                          pair<ll,ll>,vector<pair<ll,ll> >,greater<pair<ll,ll> >
#define                          mod                         1000000007
vector<ll>vt[mx];
bool vis[mx],vertex,edge;
ll coun=0;
ll dfs(ll u){
 coun++;
   for(auto x:vt[u]){
     if(!vis[x]){

            vis[x]=1;
        dfs(x);
     }
   }
return coun;
}

int main(){

     ll i,n,j,ans=0,m,s=0,sum=0,x=0,u,v;
     in2(vertex,edge);
     for(i=0;i<edge;i++){
        in2(u,v);
        vt[u].pb(v);
        vt[v].pb(u);

     }

     for(i=1;i<=vertex;i++){
         ll cnt=0;coun=0;
         if(!vis[i]){
           cnt= dfs(i);
         }
         ans=max(ans,cnt);
         coun=0,cnt=0;
     }
     ans--;
     out(ans);


}