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)

/*
  NAME : RISHABH SINGH
  INSTITUTION : DELHI TECHNOLOGICAL UNIVERSITY
*/
#include "bits/stdc++.h"
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define int long long
#define pii pair<int, int>
#define bi __builtin_inf()
#define ll long long
#define mp make_pair
#define ss second
#define pb push_back
#define ff first
#define speed_cubing ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
#define um unordered_map
#define us unordered_set
#define bp __builtin_popcount
#define ld long double
#define rep(n) for (int i = 0; i < n; i++)
#define pll pair<ll, ll>
#define mx = 1e18;
#define lb(v, val) (lower_bound(v.begin(), v.end(), val) - v.begin())
#define ub(v, val) (upper_bound(v.begin(), v.end(), val) - v.begin())
#define PI 3.1415926535
typedef tree<pii, null_type, less<pii>, rb_tree_tag, tree_order_statistics_node_update> pbds;
ll powmod(ll base,ll exp,ll MOD){ll res=1;while(exp>0){if(exp%2==1) res=(res*base)%MOD;base=(base*base)%MOD;exp/=2;}return (res%MOD);}
#define sz(v) (int)v.size()
#define endl '\n'
const int mod = 1e9+7;
ll mul_inv(ll a,ll b = mod) {ll t1=a,t2=b,t3,v1=1,v2=0,v3;while(t2!=1){ll x=t1/t2;t3=t1-x*t2;v3=v1-x*v2;t1=t2,t2=t3;v1=v2,v2=v3;}return(v2+b)%b;}
const int N = 2e5+5;
int p[N];

int fin(int i){
    return p[i]==i ? i : p[i] = fin(p[i]);
}

void uni(int i,int j){
     
        int a = fin(i),b = fin(j);
        p[a] = b;

}
 
int32_t main(){
   
   speed_cubing
   
   
   int n;
   cin>>n;
   int m;
   cin>>m;

   for(int i=1;i<=n;i++)
      p[i] = i;

   for(int i=0;i<m;i++){
      
      int x,y;
      cin>>x>>y;
      uni(x,y);

   }

   map<int,int> mm;
   int maxi = 0;

   for(int i=1;i<=n;i++)
      mm[fin(i)]++;
  
    for(auto x:mm)
       maxi = max(maxi,x.ss);

    cout<<maxi;   


   return 0;
}