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 rep(i, n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int, int>;

struct UnionFind
{
	vector<int> par;
	vector<int> rank;
	vector<int> count;

	UnionFind(int n) : par(n), rank(n), count(n){
		rep(i, n)
		{
			par[i] = i;
			rank[i] = 0;
			count[i] = 0;
		}
	}

	int find(int x){
		if (par[x] == x){
			return x;
		}
		else{
			return par[x] = find(par[x]);
		}
	}

	void unite(int x, int y){
		x = find(x);
		y = find(y);
		if (x == y)
			return;

		if (rank[x] < rank[y]){
			par[x] = y;
		}
		else{
			par[y] = x;
			if (rank[x] == rank[y])
				rank[x]++;
		}
	}

	bool same(int x, int y){
		return find(x) == find(y);
	}

	void body()
	{
		rep(i, par.size()){
			cout << par[i] << " " << rank[i] << endl;
		}
	}
	int cnt()
	{
		int ma = 0;
		rep(i, par.size()){
			count[par[i]]++;
		}
		rep(i, par.size()){
			ma = max(ma, count[par[i]]);
		}
		return (ma);
	}
};

int main()
{
	int n, m;
	cin >> n >> m;
	UnionFind uni(n);
	vector<P> p(m);
	rep(i, m){
		cin >> p[i].first >> p[i].second;
	}
	sort(p.begin(), p.end());
	reverse(p.begin(),p.end());
	rep(i, m){
		uni.unite(p[i].first - 1, p[i].second - 1);
	}
	cout << uni.cnt() << endl;
}