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# (.NET Core 3.1.201) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.Globalization;
using System.IO.Pipes;
using System.Runtime;
 
namespace Csharponly
{
    class Program
    {
        static public void Main()
        {
        	int[] NM = Arrayint();
        	int[] N = new int[NM[0]];
        	int[] H = Arrayint();
        	for(int i = 0; i < NM[1]; i++)
        	{               
        	    int[] AB = Arrayint();
        	    if(H[AB[0]-1] >= H[AB[1]-1])
        	    {
        	        N[AB[1]-1]++;
        	    }
        	    if(H[AB[0]-1] <= H[AB[1]-1])
        	    {
        	        N[AB[0]-1]++;
        	    }
        	}
			
        	int ans = 0;
        	for(int i = 0; i < NM[0]; i++)
        	{
        	    if(N[i] == 0)
        	    {
        	        ans++;
        	    }
        	}
			Console.WriteLine(ans); 
        }
 
　　　　//一つの文字列を受け渡す
        static string Onestring()
        {
            return Console.ReadLine();
        }
 
        //複数の文字列を配列で受け渡す
        //文字列と文字列の間は" "で区切る
        static string[] Arraystring()
        {
            return Console.ReadLine().Split(' ');
        }
 
        //一つの数字を受け渡す
        static int Oneint()
        {
            return int.Parse(Console.ReadLine());
        }
 
        //複数の数字を配列で受け渡す
        //数字と数字の間は" "で区切る
        static int[] Arrayint()
        {
            String[] input_line = Console.ReadLine().Split(' ');
            int[] input_num = new int[input_line.Length];
            for (int i = 0; i < input_line.Length; i++)
            {
                input_num[i] = int.Parse(input_line[i]);
            }
            return input_num;
        }
    }
}