Problem:
Score : 300 points

Problem Statement

There are N observatories in AtCoder Hill, called Obs. 1, Obs. 2, ..., Obs. N. The elevation of Obs. i is H_i.
There are also M roads, each connecting two different observatories. Road j connects Obs. A_j and Obs. B_j.

Obs. i is said to be good when its elevation is higher than those of all observatories that can be reached from Obs. i using just one road.
Note that Obs. i is also good when no observatory can be reached from Obs. i using just one road.

How many good observatories are there?

Constraints

2 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq H_i \leq 10^9

1 \leq A_i,B_i \leq N

A_i \neq B_i

Multiple roads may connect the same pair of observatories.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
H_1 H_2 ... H_N
A_1 B_1
A_2 B_2
:
A_M B_M

Output

Print the number of good observatories.

Sample Input 1

4 3
1 2 3 4
1 3
2 3
2 4

Sample Output 1

2

From Obs. 1, you can reach Obs. 3 using just one road. The elevation of Obs. 1 is not higher than that of Obs. 3, so Obs. 1 is not good.

From Obs. 2, you can reach Obs. 3 and 4 using just one road. The elevation of Obs. 2 is not higher than that of Obs. 3, so Obs. 2 is not good.

From Obs. 3, you can reach Obs. 1 and 2 using just one road. The elevation of Obs. 3 is higher than those of Obs. 1 and 2, so Obs. 3 is good.

From Obs. 4, you can reach Obs. 2 using just one road. The elevation of Obs. 4 is higher than that of Obs. 2, so Obs. 4 is good.

Thus, the good observatories are Obs. 3 and 4, so there are two good observatories.

Sample Input 2

6 5
8 6 9 1 2 1
1 3
4 2
4 3
4 6
4 6

Sample Output 2

3

Solution in Scala (2.13.1) (judge verdict: Accepted)

import java.util.Scanner
import scala.io.StdIn._

object Main extends App {

  val Array(n, m) = readLine().split(" ").map(_.toInt)
  val H = readLine().split(" ").map(_.toInt)
  val A = Array.fill(m)(readLine().split(" ").map(_.toInt - 1))


  def solve() = {
    val default = Map.empty[Int, Seq[Int]]
    val hAndRootMap = A.foldLeft(default) { (acc, root) =>
      val s = root(0)
      val e = root(1)
      acc
        .updated(s, e +: acc.getOrElse(s, Seq.empty))
        .updated(e, s +: acc.getOrElse(e, Seq.empty))
    }
    H.zipWithIndex.foldLeft(0L) { case (acc, (h, index)) =>
      hAndRootMap.get(index) match {
        case None => acc+1
        case Some(list) => if(list.map(H).forall(_ < h)) acc+1 else acc
      }
    }
  }

  println(solve())

}