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 <bits/stdc++.h>
#define rep(i, n) for(int i = 0; i < (n); i++)
using namespace std;
using ll = long long;
using pii = pair<int, int>;

struct UnionFind {
	vector<int> p;
	vector<int> s;
	
	UnionFind(int n) : p(n), s(n) {
		rep(i, n) p[i] = i;
		rep(i, n) s[i] = 1;
	}
	
	int root(int x) {
		return (p[x] == x) ? x : p[x] = root(p[x]);
	}
	
	void unite(int x, int y) {
		x = root(x);
		y = root(y);
		if (x == y) return;
		if (s[x] < s[y]) swap(x, y);
		s[x] += s[y];
		p[y] = x;
	}
	
	bool same(int x, int y) {
		return root(x) == root(y);
	}
	
	int size(int x) {
		return s[root(x)];
	}
};

int main() {
	int n, m;
	cin >> n >> m;
	UnionFind uf(n);
	rep(i, m) {
		int a, b;
		cin >> a >> b;
		a--, b--;
		uf.unite(a, b);
	}
	int ans = 0;
	rep(i, n) ans = max(ans, uf.size(i));
	cout << ans << endl;
	return 0;
}