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)

// ABC177D - Friends

#include <bits/stdc++.h>
using namespace std;

typedef unsigned long long int ull;
typedef long long int ll;

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

	// n要素で初期化
	UnionFind(int n) {
		par.resize(n);
		rank.resize(n);
		count.resize(n);
		for (int i = 0; i < n; i++) {
			par[i] = i;
			rank[i] = 0;
			count[i] = 1;
		}
	}

	// 木の根を求める。
	int find(int x) {
		if (par[x] == x) {
			return x;
		} else {
			return par[x] = find(par[x]);
		}
	}

	// xとyの属する集合を併合
	void unite(int x, int y) {
		x = find(x);
		y = find(y);
		if (x == y) {
			return;
		}
		
		count[x] += count[y];
		count[y] = count[x];
		if (rank[x] < rank[y]) {
			par[x] = y;
		} else {
			par[y] = x;
			if (rank[x] == rank[y]) {
				rank[x]++;
			}
		}
	}

	// xとyが同じ集合に属するか否か
	bool same(int x, int y) {
		return find(x) == find(y);
	}

	// xと同じ集合の要素数を返す。
	int getCount(int x) {
		return count[x];
	}
};

int main() {
	int N, M;
	cin >> N >> M;

	UnionFind tree(N);
	for (int i = 0; i < M; i++) {
		int A, B;
		cin >> A >> B;
		tree.unite(A-1, B-1);
	}

	int ans = 0;
	for (int i = 0; i < N; i++) {
		ans = max(ans, tree.getCount(i));
	}

	cout << ans << endl;

	return 0;
}