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>
using namespace std;
#define rep(i, n) for (ll i = 0; i < n; i++)
#define repl(i, l, r) for (ll i = (l); i < (r); i++)
#define all(x) (x).begin(), (x).end()
using ll = long long;
const ll MOD = 1000000007;
struct UnionFind {
	vector<int> par;
	UnionFind(int n) : par(n, -1) {}
	void init(int n) { par.assign(n, -1); }
	int root(int x) {
		if (par[x] < 0) return x;
		return par[x] = root(par[x]);
	}
	bool issame(int x, int y) { return root(x) == root(y); }
	bool merge(int x, int y) {
		x = root(x), y = root(y);
		if (x == y) return false;
		if (par[x] > par[y]) swap(x, y);
		par[x] += par[y];
		par[y] = x;
		return true;
	}
	int size(int x) { return -par[root(x)]; }
};
int main() {
	int n, q;
	scanf("%d%d", &n, &q);
	UnionFind tree(n);
	while (q--) {
		int a, b, c;
		scanf("%d%d", &a, &b);
		a--, b--;
		tree.merge(a, b);
	}
	int ans = 0;
	for (auto t : tree.par) ans = min(ans, t);
	ans *= -1;
	printf("%d\n", ans);
	return 0;
}