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: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;

const int N = 2 * 1e5 + 10;
const int M = 2 * 1e5 + 10;
int n, m, maxn;
int s[N], f[N], fa[N];

int get_fa (int x) {
	if (fa[x] == x) return x;
	else return fa[x] = get_fa (fa[x]);
}

int main () {
	scanf ("%d%d", &n, &m);
	for (int i = 1; i <= n; i ++ ) fa[i] = i;
	for (int i = 1; i <= m; i ++ ) {
		int x, y, fx, fy;
		scanf ("%d%d", &x, &y);
		fx = get_fa (x);
		fy = get_fa (y);
		if (fx == fy) continue;
		else {
			if (!f[x]) s[fx] ++;
			if (!f[y]) s[fy] ++;
			f[x] = f[y] = 1;
			fa[fx] = fy;
			s[fy] += s[fx];
			s[fx] = 0;
		}
	}
	for (int i = 1; i <= n; i ++ ) maxn = max (maxn, s[i]);
	printf ("%d\n", maxn);
	return 0;
}