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;

int n, m;
vector<vector<int>> edges;
vector<bool> v;

void dfs(int ver, vector<int> &com) {
	v[ver] = 1;
	com.push_back(ver);
	for (int nb : edges[ver]) {
		if (!v[nb]) dfs(nb, com);
	}
}

int main()
{
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	
	cin >> n >> m;
	edges.resize(n + 1);
	v.resize(n + 1);
	for (int i = 0; i < m; i++) {
		int x, y;
		cin >> x >> y;
		edges[x].push_back(y);
		edges[y].push_back(x);
	}
	
	int ans{};
	for (int i = 1; i <= n; i++) {
		vector<int> c;
		if (!v[i]) dfs(i, c);
		ans = max(ans, (int) c.size());
	}
	cout << ans;
}