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;
typedef long long ll;
vector<int>e[200005];
int use[200005];
int num[200005];
int n, m, cnt;
void bfs(int x) {
	if (use[x])return;
	use[x] = cnt;
	for (int i = 0; i < e[x].size(); ++i)bfs(e[x][i]);
}
int main() {
	cin >> n >> m;
	for (int i = 0; i < m; ++i) {
		int x, y;
		scanf("%d%d", &x, &y);
		e[x].push_back(y);
		e[y].push_back(x);
	}
	for (int i = 1; i <= n; ++i) {
		if (!use[i]) {
			++cnt;
			bfs(i);
		}
	}
	for (int i = 1; i <= n; ++i) {
		++num[use[i]];
	}
	int mx = 1;
	for (int i = 1; i <= n; ++i) {
		mx = max(mx, num[i]);
	}
	cout << mx << endl;
	return 0;
}