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.Linq;

namespace AtCoder
{
    class Program
    {
        static void Main(string[] args)
        {
            string[] input = Console.ReadLine().Split(' ');
            int N = int.Parse(input[0]);
            int M = int.Parse(input[1]);
            int[] observatory = Array.ConvertAll(Console.ReadLine().Split(' '), int.Parse);
            int[] good = Array.ConvertAll(observatory, x => x - x);

            for(int i= 0; i < M; i++)
            {
                int[] load = Array.ConvertAll(Console.ReadLine().Split(' '),int.Parse);
                int to = load[0];
                int from = load[1];
                if (observatory[to - 1] > observatory[from - 1])
                {
                    good[from - 1] += 1;
                }
                else if(observatory[to - 1] < observatory[from - 1])
                {
                    good[to - 1] += 1;
                }
                else
                {
                    good[from - 1] += 1;
                    good[to - 1] += 1;
                }
            }

            Console.WriteLine(Array.FindAll(good,x=> x==0).Length);
        }
    }
}