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>
#define endl '\n'
#define F first
#define S second
#define PB push_back
#define MP make_pair
#define FASTIO ios::sync_with_stdio(0); cin.tie(0);
#define REP(i,a,b) for(int i=a;i<b;i++)
#define mod 1000000007
const int inf=2e3;
#define maxx 200005
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> pi;
typedef pair<ll,ll> pll;
vi adj[maxx];
int cnt;
bool vis[maxx];
void dfs(int src){
   vis[src]=1;
   cnt++;
   for(int i=0;i<adj[src].size();i++){
    if(!vis[adj[src][i]]){
        dfs(adj[src][i]);
    }
   }
   return;
}
int main(){
    FASTIO;
    int n,m,a,b,mx=0;
    cin>>n>>m;
    REP(i,0,m){
      cin>>a>>b;
      adj[a].PB(b);
      adj[b].PB(a);
    }
    REP(i,1,n+1){
      cnt=0;
      if(!vis[i]){
        dfs(i);
        mx=max(mx,cnt);
      }
    }
    cout<<mx<<endl;
    return 0;
}