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 C# (Mono-csc 3.5.0) (judge verdict: Accepted)

using System;
using System.Linq;
using System.Collections.Generic;

public static class Program
{
  public static void Main(string[] args)
  {
    string[] input1=Console.ReadLine().Split(' ');
    int N=int.Parse(input1[0]);
    int M=int.Parse(input1[1]);
    int[] Observatory=Console.ReadLine().Split(' ').Select(x=>int.Parse(x)).ToArray();
    int[] ma=new int[N];
    


    for(int i=0;i<M;i++) {
        int[] Path=Console.ReadLine().Split(' ').Select(x=>int.Parse(x)).ToArray();
        ma[Path[0]-1]=Math.Max(ma[Path[0]-1],Observatory[Path[1]-1]);
        ma[Path[1]-1]=Math.Max(ma[Path[1]-1],Observatory[Path[0]-1]);
    }
    
    
    int count=0;
    for(int i=0;i<N;i++) {
        if(ma[i]<Observatory[i]) {
            count++;
        }
    }
 
    Console.WriteLine(count);
 
  }


}