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 <iostream>
#include <map>
#include <cmath>
#include <queue>
#include <vector>
#include <string>
#include <algorithm>
#include <functional>
using namespace std;
#pragma warning (disable: 4996)

int N, M, A[1 << 18], B[1 << 18];
int cnt = 0;
bool used[1 << 18];
vector<int> X[1 << 18];

void dfs(int pos) {
	if (used[pos] == true) return;
	used[pos] = true;
	cnt++;
	for (int i : X[pos]) {
		if (used[i] == true) continue;
		dfs(i);
	}
}

int main() {
	cin >> N >> M;
	for (int i = 1; i <= M; i++) {
		cin >> A[i] >> B[i];
		X[A[i]].push_back(B[i]);
		X[B[i]].push_back(A[i]);
	}

	int Answer = 0;
	for (int i = 1; i <= N; i++) {
		if (used[i] == true) continue;
		cnt = 0;
		dfs(i);
		Answer = max(Answer, cnt);
	}
	cout << Answer << endl;
	return 0;
}