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: Wrong Answer)

import kotlin.comparisons.compareBy

fun main(args: Array<String>){
    val (n,m) = nextIntList()
    val Hs = nextIntList()
    val tenboRes = arrayOfNulls<Boolean>(n)

    var As = Array(m) {IntArray(2)}
    for(i in 0 until m){
        val a= nextIntList()
        As[i] = intArrayOf(a[0],a[1])
    }

    As.forEachIndexed { index, ints ->
        if(Hs[ints[0] - 1] > Hs[ints[1] - 1]) As[index] = intArrayOf(ints[1],ints[0])
    }

    val ckInts = IntArray(2)
    As = As.sortedWith(compareBy {
        it[0]
    } ).mapNotNull {
        if(ckInts[0]==0){
            ckInts[0] = it[0]
            ckInts[1] = it[1]
            it
        }else if(ckInts[0] == it[0] && ckInts[1] == it[1]){
            null
        }else{
            it
        }
    }.toTypedArray()




    As.forEach { println("${it[0]} , ${it[1]}") }

    As.forEach {
        if(it[0] == it[1]){
            tenboRes[it[0]-1] = false
            tenboRes[it[1]-1] = false
        }else{
            tenboRes[it[0]-1] = check(tenboRes[it[0]-1])
            tenboRes[it[1]-1] = false
        }


    }

    val res = tenboRes.count { it == null || it}

    println(res)
}
fun next() = readLine()!!
fun nextIntList() = next().split(" ").map{ it.toInt() }

fun check(b:Boolean?):Boolean{
    b?: return true
    return if(!b) b else true
}