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)

#include <bits/stdc++.h>
using namespace std;
//#define int long long
#define pb push_back
#define mp make_pair
#define F first
#define S second

signed main(){
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif

	int n,m;
	cin >> n >> m;
	vector <int> v[n+1];
	int a,b;
	for(int i=0; i<m; i++){
		cin >> a >> b;
		v[a].pb(b);
		v[b].pb(a);
	}

	int vis[n+1] = {};
	int g=0,z=0;
	stack <int> s;
	//q.push(1);

	for(int i=1; i<=n; i++){
		if(vis[i]==1)
			continue;

		g=0;
		vis[i]=1;
		s.push(1);
		while(!s.empty()){
			int x = s.top();
			s.pop();
			g++;
			for(int j=0; j<v[x].size(); j++){
				int a = v[x][j];
				if(vis[a]==1)
					continue;

				vis[a]=1;
				s.push(a);
			}
		}
		z = max(g,z);
	}

	cout << z;




	
	

	return 0;
}