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;
const int mod = 1e9 + 7;
const int N = 200006;
vector<int> E[N];
vector<int> seen(N, false);
int cnt = 0;

void dfs(int node) {
	seen[node] = true;
	cnt += 1;
	for(auto it : E[node]) {
		if(!seen[it]) {
			dfs(it);
		}
	}
}

int main() {
	
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	
	int n, m;
	cin >> n >> m;
	for(int i = 1; i <= m; i++) {
		int x, y;
		cin >> x >> y;
		E[x].push_back(y);
		E[y].push_back(x);
	}
	
	vector<int> sz;
	
	for(int i = 1; i <= n; i++) {
		if(!seen[i]) {
			cnt = 0;
			dfs(i);
			sz.push_back(cnt);
			
		}
	}
	
	sort(sz.begin(), sz.end());
	int ans = 0;
	int rem = 0;
	
	for(auto &it : sz) {
		it -= rem;
		ans += it;
		rem += it;
	}
	
	cout << ans << '\n';
	
	return 0;
}