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 Scala (2.13.1) (judge verdict: Wrong Answer)

import scala.io.StdIn._

object Main extends App {

  val Array(n,m) = readLine().split(" ").map(_.toInt)
  val AB = Array.fill(m)(readLine().split(" ").map(_.toInt-1))
    .map(list => (list(0), list(1)))
  def solve() = {
    val linkMap = (0 until n).map(i => (i -> Set(i))).toMap

    AB.foldLeft((linkMap, 0)) { case ((accMap, maxLink), (a,b)) =>
      val setA = accMap(a)+b
      val setB = accMap(b)+a
      val updatedMap = accMap.updated(a, setA).updated(b, setB)
      val tmpMaxLink = Math.max(setA.size, setB.size)
      (updatedMap, Math.max(maxLink, tmpMaxLink))
    }._2
  }

  println(solve())

}