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>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <vector>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <string>

#define FastIO 					\
	ios::sync_with_stdio(0); 	\
	cin.tie(NULL); 				\
	cout.tie(NULL);

using namespace std;

typedef long long ll;

void solve() {
	int n, m;
	cin >> n >> m;
	int a[m], b[m];
	for (int i = 0; i < m; i++) {
		cin >> a[i] >> b[i];
	}
	map<int, unordered_set<int>> mp;
	for (int i = 1; i <= n; i++) {
		unordered_set<int> us;
		us.insert(i);
		mp.insert({i, us});
	}
	for (int i = 0; i < m; i++) {
		mp[a[i]].insert(b[i]);
		mp[b[i]].insert(a[i]);
	}
	int ans = 0;
	// cout << "\n";
	for (auto x : mp) {
		unordered_set<int> ts = x.second;
		/*
		set<int> ss;
		ss.insert(ts.begin(), ts.end());
		for (auto sx : ss) {
			cout << sx << " ";
		}
		cout << "\n";
		*/
		if (ts.size() > ans) {
			ans = ts.size();
		}
	}
	// cout << "\n";
	cout << ans << "\n";
}

int main() {
	FastIO;
	int n = 1;
	// cin >> n;
	while (n--) {
		solve();
	}
}