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: Wrong Answer)

object Main extends App {
  import scala.io.StdIn._
  
  val Array(n,m) = readLine.split(" ").map{_.toInt}.toArray
  val h = readLine.split(" ").map{_.toLong}.toList
  val ab = (for (i <- 1 to m) yield {
    val Array(a,b) = readLine.split(" ").map{_.toInt}.toArray
    (a,b)
  }).toList
  
  val sortedH = h.zipWithIndex.sortWith((x,y) => x._1 > y._1).map{case (x,y) => (x,y+1)}
  val check = Array.fill(n+1)(false)
  check(0) = true
  var ans = 0
  
  for (x <- sortedH if check.count{x => !x} != 0) {
    val i = x._2
    if (!check(i)) {
      ans += 1
      check(i) = true
    }
    ab.filter{case (a,b) => a == i || b == i}.map{case (a,b) => if (a == i) b else a}.foreach{i => check(i) = true}
  }
 
  println(ans)
                                         
}