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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import static java.lang.Integer.*;
import static java.lang.Long.*;
import static java.lang.System.*;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Main {

    static void solve() throws Exception {
        int N = scanInt(), M = scanInt();
        long[] ten = new long[N];
        int[] ans = new int[N];
        for (int i = 0; i < N; i++) {
            ten[i] = scanLong();
        }
        for (int i = 0; i < M; i++) {
            int A =  scanInt() - 1;
            int B =  scanInt() - 1;
            if (ten[A] < ten[B]) {
                ans[A] = 1;
            } else if (ten[A] > ten[B]) {
                ans[B] = 1;
            } else {
                ans[A] = 1;
                ans[B] = 1;
            }
        }
        int count = 0;
        for (int i = 0; i < N; i++) {
            if (ans[i] == 0) count++;
        }
        out.print(count);
    }

    static int scanInt() throws IOException {
        return parseInt(scanString());
    }

    static long scanLong() throws IOException {
        return parseLong(scanString());
    }

    static String scanString() throws IOException {
        while (tok == null || !tok.hasMoreTokens()) {
            tok = new StringTokenizer(in.readLine());
        }
        return tok.nextToken();
    }

    static BufferedReader in;
    static PrintWriter out;
    static StringTokenizer tok;

    public static void main(String[] args) {
        try {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
            solve();
            in.close();
            out.close();
        }
        catch (Throwable e) {
            e.printStackTrace();
            exit(1);
        }
    }
}