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: Accepted)

object Main extends App {
  val sc = new java.util.Scanner(System.in)
  val n, m = sc.nextInt()
  val uft = new UnionFindTree(n)

  for (_ <- 1 to m) {
    val a, b = sc.nextInt() - 1
    uft.unite(a, b)
  }

  val ans = (0 until n).map(uft.size).max
  println(ans)
}

class UnionFindTree(n: Int) {
  private val par = Array.tabulate(n)(identity)
  private val rank = Array.fill(n)(1)
  private val size0 = Array.fill(n)(1)

  def root(x: Int): Int = {
    var res = x
    var children = List[Int]()
    while (par(res) != res) {
      children ::= res
      res = par(res)
    }

    for (y <- children) {
      par(y) = res
    }
    res
  }

  def unite(x: Int, y: Int): Unit = {
    val xr = root(x)
    val yr = root(y)
    if (xr != yr) {
      if (rank(xr) < rank(yr)) {
        par(xr) = yr
        size0(yr) += size0(xr)
      } else {
        par(yr) = xr
        size0(xr) += size0(yr)
        if (rank(xr) == rank(yr)) {
          rank(xr) += 1
        }
      }
    }
  }

  def size(x: Int): Int = size0(root(x))
}