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 PHP (7.4.4) (judge verdict: Wrong Answer)

<?php
fscanf(STDIN, "%d %d", $N, $M);
$r = [];
$g = array_fill(1, $N, null);
$ans = 0;
$cnt = 0;
for ($i = 0; $i < $M; $i++) {
    fscanf(STDIN, "%d %d", $A, $B);
    $r[$A][$B] = 1;
    $r[$B][$A] = 1;
}
for ($i = 1; $i <= $N; $i++) {
    if (is_null($g[$i])) {
        $cnt++;
        $g[$i] = $cnt;
    }
    $memo = [];
    $stack = [];
    while(true) {
        if (!isset($r[$i])) break;
        foreach ($r[$i] as $k => $v) {
            if (!in_array($k, $memo)) {
                $memo[] = $k;
                $stack[] = $k;
            }
        }
        if (!$stack) break;
        while($stack) {
            $x = array_pop($stack);
            if (is_null($g[$x])) {
                $g[$x] = $cnt;
            }
        }
    }
}
// var_dump(array_count_values($g));
printf("%d", max(array_count_values($g)));