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>
#define INF 1e9
using namespace std;

//자동 정렬은 ctrl k f
//주석은 ctrl k c
//주석 해제는 ctrl k u
// int gcd(int a, int b){
// 	while(b!=0){
// 		int r= a%b;
// 		a=b;
// 		b=r;
// 	}
// 	return a;
// }
// int lcm(int a, int b){
// 	return a*b/gcd(a,b);
// }
int arr[200001];
int save[200001];

int getParent(int x) {
	if (arr[x] == x) {
		return x;
	}
	else {
		return arr[x] = getParent(arr[x]);
	}

}

void find(int x, int y) {
	//int x_p = arr[x];
	int x_p = getParent(x);
	//int y_p = arr[y];
	int y_p = getParent(y);

	if (x_p != y_p) {
		if (x_p < y_p) {
			arr[y_p] = x_p;
		}
		else {
			arr[x_p] = y_p;
		}
	}
}

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);

	int n, m;
	cin >> n >> m;

	for (int i = 1; i <= n; i++) {
		arr[i] = i;
	}

	
	for (int i = 0; i < m; i++) {
		int x, y;
		cin >> x >> y;
		find(x, y);

	}
	
	for (int i = 1; i <= n; i++) {
		save[arr[i]]++;
	}

	int ans = 0;
	for (int i = 1; i <= n; i++) {
		ans = max(ans, save[i]);
	}
	cout << ans;
	return 0;
}