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 Kotlin (1.3.71) (judge verdict: Runtime Error)

fun main(args : Array<String>){
    var (n,m)=readLine()!!.split(" ").map(String::toInt)
    var h = readLine()!!.split(" ").map(String::toInt)

    var patharray = ArrayList<Pair<Int, Int>>()
    for(i in 1..m){
        var (a,b)=readLine()!!.split(" ").map(String::toInt)
        patharray.add(Pair(a,b))
    }

    var ans = 0
    for((hindex,hheight) in h.withIndex()){
        var f: Boolean = true
        
        var thispath = patharray.filter { 
            it.first == hindex 
        }
        for(it in thispath){
            if(h[it.second] >= hheight){
                f = false
                break
            }
        }

        if(f){
            var thispath2 = patharray.filter { 
                it.second == hindex 
            }
            for(it in thispath2){
                if(h[it.first] >= hheight){
                    f =false
                    break
                }
            }
        }

        if(f == true) ans++
    }
  print(ans.toString())
}