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>
using namespace std;
typedef long long ll;
const int MOD = 1e9+7;
int find(vector<int>& ds,int i) {
	while(ds[i]!=i) {
		i = ds[i];
	}
	return i;
}
void combine(vector<int>& ds,vector<int>& p,int u,int v) {
	int x = find(ds,u);
	int y = find(ds,v);
	if(x!=y) {
		ds[x] = min(x,y);
		ds[y] = min(x,y);
		p[min(x,y)] += p[max(x,y)];
	}
}
void solve() {
	int n,m;
	cin>>n>>m;
	vector<int> ds(n+1,0),p(n+1,0);
	for(int i=0;i<=n;i++) {
		ds[i] = i;
		p[i] = 1;
	}
	while(m--) {
		int u,v;
		cin>>u>>v;
		combine(ds,p,u,v);	
	}
	m = p[1];
	for(int i=2;i<=n;i++) {
		m = max(m,p[i]);
	}
	cout<<m;
}
int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	int test = 1;
	while(test--) {
		solve();
	}
}