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: Accepted)


var t = "6 5\n" +
        "8 6 9 1 2 1\n" +
        "1 3\n" +
        "4 2\n" +
        "4 3\n" +
        "4 6\n" +
        "4 6"

fun main(args: Array<String>) {

    read = stdin

    val (N, M) = readInts()
    val H = readInts()
    val (A, B) = readInts2(M)

    val flags = Array(N){true}

    for(i in 0 until M){
        A[i]--
        B[i]--
        if(H[A[i]] > H[B[i]]){
            flags[B[i]] = false
        }else if(H[A[i]] == H[B[i]]){
            flags[A[i]] = false
            flags[B[i]] = false
        }else{
            flags[A[i]] = false
        }
    }

    var cnt = 0
    for(flag in flags) {
        if(flag) cnt++
    }
    println(cnt)
}


fun answer(args: Array<String>){
    println("ほげ")
}



// #######################################################
//  入力用
// #######################################################

var ts = t.split("\n")
var c = 0
var debug = {ts[c++]}
var stdin = {readLine()!!}
var read = stdin

fun readInt(): Int{
    return read().toInt()
}

fun readInts(): Array<Int>{
    return read().split(" ").map(String::toInt).toTypedArray()
}

fun readInts2(n: Int) : Pair<Array<Int>, Array<Int>>{
    val a = Array(n){0}
    val b = Array(n){0}
    for(i in 0 until n){
        val r = readInts()
        a[i] = r[0]
        b[i] = r[1]
    }
    return Pair(a, b)
}