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 java.util.*

fun main(args: Array<String>) {
    val sc = Scanner(System.`in`)
    val n = sc.nextInt()
    val m = sc.nextInt()
    val h = (0 until n).map { it + 1 to sc.next().toLong() }
    val ab = (0 until m).map { sc.next().toInt() to sc.next().toInt() }
    println(problem166c(n, m, h, ab))
}

fun problem166c(
    n: Int,
    m: Int,
    h: List<Pair<Int, Long>>,
    ab: List<Pair<Int, Int>>
): Int {
    val heights = h.sortedBy { it.second }.toMap()
    var count = 0
    val routes = hashMapOf<Int, Long>()
    for (i in 0 until m) {
        val abi = ab[i]
        if (routes.containsKey(abi.first)) {
            routes[abi.first]!!.plus(Math.max(routes[abi.first]!!, heights[abi.second]!!))
        } else {
            routes[abi.first] = heights[abi.second]!!
        }
        if (routes.containsKey(abi.second)) {
            routes[abi.second]!!.plus(Math.max(routes[abi.second]!!, heights[abi.first]!!))
        } else {
            routes[abi.second] = heights[abi.first]!!
        }
    }
    for (i in 1..n) {
        if (!routes.containsKey(i)) {
            count++
            continue
        }
        if (routes[i]!! < heights[i]!!) count++
    }
    return count
}