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.io._
import java.util.StringTokenizer
import scala.util.control.Breaks.{breakable,break}
import scala.collection.mutable.{Map, Stack, Queue, PriorityQueue, Set}

object Main extends App {
  val in = new InputReader(System.in)
  val pw = new PrintWriter(System.out)

  // val n = in.next().toInt
  val N, M = in.next().toInt
  val _H = new Array[(Int, Int)](N)
  val high = new Array[Int](N+1)
  for (i <- 0 until N) {
    val h = in.next().toInt
    _H(i) = (h, i+1)
    high(i+1) = h
  }

  val graph = Array.fill[List[Int]](N+1)(Nil)
  for (i <- 1 to M) {
    val A, B = in.next().toInt
    graph(A) = B :: graph(A)
    graph(B) = A :: graph(B)
  }

  val H = _H.sorted.reverse

  // 
  val dp = new Array[Boolean](N+1)
  var ans = 0
  for ((h, node) <- H) {
    if (!dp(node)) {
      ans += 1

      dp(node) = true

      var flag = false
      for (e <- graph(node)) {
        if (!dp(e))
          dp(e) = true

        // 同一 h のチェック
        if (high(e) >= h)
          flag = true
      }

      if (flag)
        ans -= 1
    }
  }

  println(ans)

}

object Sub {
  def print2DimArray[A](mat: Array[Array[A]]): Unit = {
    mat.foreach(lin => println(lin.toList))
  }
}

class InputReader(stream: InputStream) {
  private[this] val reader = new BufferedReader(new InputStreamReader(stream))
  private[this] var tokenizer: StringTokenizer = new StringTokenizer(reader.readLine())
  def next(): String = {
    while (!tokenizer.hasMoreTokens()) {
      tokenizer = new StringTokenizer(reader.readLine())
    }
    tokenizer.nextToken()
  }
}