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);
$friend_r = array();
$log = array();
$que = array();
for($i=1;$i<=$n;$i++) $log[$i] = -1;
for($i=0;$i<$m;$i++){
  fscanf(STDIN,"%d%d",$a,$b);
  $friend_r[$a][] = $b;
  $friend_r[$b][] = $a;
}
//print_r($friend_r);

$count = 0;
$que[] = 1;
$log[1] = 1;
for($i=1;$i<$n;$i++){
  if($log[$i]==1){
    continue;
  }
  while(count($que)>0){	
    $person = array_shift($que);
    if(!isset($friend_r[$person])){
      continue;
    }
    foreach($friend_r[$person] as $friend){
      //echo $friend."\n";
      if($log[$friend]==-1){
          $que[] = $friend;
          $log[$friend] = 1;
          $count++;
      }
    }
   // print_r($que);
  }
}

echo ($count+1);

//echo ($max+1);