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: Runtime Error)

#include <stdio.h>

int	main(void)
{
	int	n, m;
	int	a, b;
	int	gs[100001] = {0};
	int	g, g_max = 0;
	int	fr[50001] = {0};
	int	max = 0;

	scanf("%d %d", &n, &m);
	for (int i = 0; i < m; i++)
	{
		g = 0;
		scanf("%d %d", &a, &b);
		if (!gs[a] && !gs[b])
		{
			g_max++;
			fr[g_max] += 2;
			g = g_max;
		}
		else if (gs[a] && gs[b])	// どちらも値がある
		{
			if (gs[a] == gs[b])
				g = gs[a];
			else if (gs[a] < gs[b])
			{
				fr[gs[b]]--;
				fr[gs[a]]++;
				g = gs[a];
			}
			else
			{
				fr[gs[b]]++;
				fr[gs[a]]--;
				g = gs[b];
			}
		}
		else	// どちらかに値がない
		{
			g = 0 < gs[a] ? gs[a] : gs[b];
			fr[g]++;
		}
		gs[a] = gs[b] = g;
	}
	for (int i = 0; i <= n; i++)
	{
		if (max < fr[i])
			max = fr[i];
	}
	printf("%d\n", max);
	return 0;
}