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: Wrong Answer)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main
{
  public static void main(String[] args) throws IOException
  {
    BufferedReader r = new BufferedReader(new InputStreamReader(System.in), 1);
    String s;
    String sl[];

    s = r.readLine();
    sl = s.split(" ");

    boolean a[] = new boolean[100001];
    int x[] = new int[100001];
    a[0] = true;

    int n = Integer.parseInt(sl[0]);
    int m = Integer.parseInt(sl[1]);

    s = r.readLine();
    sl = s.split(" ");

    for(int i = 0; i < n; i++)
    {
      x[i + 1] = Integer.parseInt(sl[i]);
    }

    for(int i = 0; i < m; i++)
    {
      s = r.readLine();
      sl = s.split(" ");
      int v1 = Integer.parseInt(sl[0]);
      int v2 = Integer.parseInt(sl[1]);
      if(v1 >= v2)
      {
        a[v2] = true;
      }
      if(v2 >= v1)
      {
        a[v2] = true;
      }
    }

    int ss = 0;
    for(int i = 1; i <= n; i++)
    {
      if(!a[i]) ss++;
    }
    System.out.println(ss);
  }

}

class RemRing
{
  long module;

  public RemRing(long module)
  {
    this.module = module;
  }

  public long sum(long a, long b)
  {
    return (a + b + module) % module;
  }

  public long sub(long a, long b)
  {
    return (a - b + module) % module;
  }

  public long prod(long a, long b)
  {
    return (a * b) % module;
  }

  public long div(long a, long b)
  {
    return (a * inv(b)) % module;
  }

  public long inv(long a)
  {
    long b = this.module;
    long u = 1;
    long v = 0;
    while (b > 0)
    {
      long t = a / b;
      a -= t * b;
      u -= t * v;
      long z = a;
      a = b;
      b = z;
      z = u;
      u = v;
      v = z;
    }
    u %= this.module;
    if (u < 0)
      u += this.module;
    return u;
  }
}