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 JavaScript (Node.js 12.16.1) (judge verdict: Accepted)

const { parse } = require("path");

function Main(input) {
    input = input.trim().split("\n").map(function(x) { return x.split(" ")});
    let N = parseInt(input[0][0], 10);
    let M = parseInt(input[0][1], 10);
    let bitVec = Array.from({length:N}, () => true);
    let H = input[1].map(e => parseInt(e, 10));
    let AB;
    for (let i = 0; i < M; i++){
        AB = input[i + 2].map(e => parseInt(e, 10));
        if (H[AB[0] - 1] > H[AB[1] - 1]){
            bitVec[AB[1] - 1] = false;
        } else if (H[AB[0] - 1] < H[AB[1] - 1]) {
            bitVec[AB[0] - 1] = false;
        } else {
            bitVec[AB[1] - 1] = false;
            bitVec[AB[0] - 1] = false;
        }
    }
    let ans = 0;
    for (let x of bitVec){
        if (x) ans++;
    }
    console.log(ans);
}

Main(require("fs").readFileSync("/dev/stdin", "utf8"));