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 <stdio.h>
#include <math.h>

typedef long long ll;
#define MAX_N (200005)

int uf[MAX_N]; // union-find
int n;

void uf_init()
{
	for (int i = 0; i < n; i++) {
		uf[i] = i;
	}
}

int uf_root(int x)
{
	if (uf[x] == x)
	{
		return x;
	}
	else
	{
		uf[x] = uf_root(uf[x]);
		return uf[x];
	}
	
}

void uf_unite(int x, int y)
{
	int rx = uf_root(x);
	int ry = uf_root(y);
	if (rx == ry)
	{
		return;
	}
	uf[rx] = ry;
}

int main() {
	int m;
	scanf("%d %d", &n, &m);
	
	uf_init();
	for (int i = 0; i < m; i++) {
		int a, b;
		scanf("%d %d", &a, &b);
		a--;
		b--;
		uf_unite(a, b);
		
	}
	
	int gr[MAX_N]; // gr[x]: root(i) = x なるiの数(友達軍団の人数)
	for (int i = 0; i < n; i++) {
		gr[i] = 0;
	}
	for (int i = 0; i < n; i++) {
		gr[uf_root(i)]++;
	}
	
	int gr_max = -1;
	for (int i = 0; i < n; i++) {
		if (gr[i] > gr_max) {
			gr_max = gr[i];
		}
	}
	printf("%d", gr_max);
	
	
	return 0;
}