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;


class friends {
private:
	int n, m;

	int ans = 0;
	int depth = 0;

	vector<set<int>> adjacencyList;
	vector<bool> seeked;

public:
	friends(int _n, int _m) {
		n = _n;
		m = _m;

		adjacencyList.resize(n + 1);
		seeked.resize(n + 1);
		for (auto e : seeked) e = false;
	}

	void setAdjList(int a, int b);
	void search(int i = 1);
	void show();
	void solve();
};

void friends::setAdjList(int a, int b) {
	adjacencyList[a].insert(b);
	adjacencyList[b].insert(a);
}

void friends::search(int i) {
	for (auto e : adjacencyList[i]) {
		if (seeked[e] == true) continue;

		depth += 1;
		seeked[e] = true;
		search(e);
	}
}

void friends::show() {
	for (int i = 1; i <= n; i++) {
		cout << i << ":";
		for (auto e : adjacencyList[i]) {
			cout << " " << e;
		}
		cout << endl;
	}

	for (int i = 1; i <= n; i++) {
		if (seeked[i] == false) cout << "未探索" << endl;;
	}
}

void friends::solve() {
	search();
	ans = depth;

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

		depth = 0;
		search(i);

		if (ans < depth) ans = depth;
	}

	cout << ans << endl;

	//show();
}

int main() {
	int n, m;
	cin >> n >> m;

	friends friends(n, m);

	int a, b;
	for (int i = 1; i <= m; i++) {
		cin >> a >> b;
		friends.setAdjList(a, b);
	}

	friends.solve();
}