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;
int par[200001],ran[200001],box[200001];
int main(void){
	int n,m,x,y,i;
	cin >> n >> m;
	for(i = 1; i <= n; i++){
		par[i] = i;
		ran[i] = 0;
		box[i] = 0;
	}
	for(i = 0; i < m; i++){
		cin >> x >> y;
		if(x < y)	swap(x,y);
		while(x != par[x])	x = par[x];
		while(y != par[y]) 	y = par[y];
		if(x != y){
			if(ran[x] >= ran[y]){
				par[y] = x;
				ran[x]++;
			}
			else	par[x] = y;
		}
	}
	for(i = 1; i <= n; i++){
		x = i;
		while(x != par[i]) x = par[i];
		box[x]++;
	}
	int MAX = INT_MIN;
	for(i = 1; i <= n; i++)	MAX = max(MAX,box[i]);
	cout << MAX <<endl;
	return 0;
}