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>
#include<algorithm>
using namespace std;
#define cool ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL) 
#define ll long long
#define MAX 1000001
#define mod 1000000007
#define pie 3.14159265358979323846
bool visited[200005]={false};
int cnt;
vector<int> adj[200005];
void DFS(int v) 
{  
    visited[v] = true;  
    for(int i=0;i<adj[v].size();i++)
	{
		if(visited[adj[v][i]]==false){
			cnt++;
			DFS(adj[v][i]);		
		}
	}
}
vector<int> fun(int N) 
{ 
	vector<int> vec;
    for (int v = 1; v <= N; v++) { 
        if (visited[v] == false) {
			cnt=1; 
            DFS(v); 
            vec.push_back(cnt);
        } 
    } 
    return vec; 
} 
 
void solve(){
	int n,m;
	cin>>n>>m;
	for(int i=0;i<m;i++){
		int x,y;
		cin>>x>>y;
		adj[x].push_back(y);
		adj[y].push_back(x);
	}
	vector<int> res;
	res=fun(n);
	cout<<*max_element(res.begin(),res.end());	
}
int main()
{ 
    cool;
    int t;
    //cin>>t;
    t=1;
    while(t--){
    	solve();
	}
    return 0;
}