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: Time Limit Exceeded)

using System;
using static AtCoderTemplate.MyLibrary;

namespace AtCoderTemplate
{
    class Program
    {
        static void Main(string[] args)
        {
            int[] input = ToIntArray(Console.ReadLine().Split(' '));
            int n = input[0], m = input[1];
            input = ToIntArray(Console.ReadLine().Split(' '));
            int[] Hs = new int[input.Length+1];
            input.CopyTo(Hs,1);
            int[][] edge = new int[Hs.Length][];
            for(int i = 0; i < edge.Length/2; i++)
            {
                edge[i] = new int[0];
                edge[Hs.Length - 1 - i] = new int[0];
            }
            edge[Hs.Length/2] = new int[0];
            int[] AB;
            int a, b;
            bool good;
            for(int i = 0;i < m;i++)
            {
                AB = ToIntArray(Console.ReadLine().Split(' '));
                a = AB[0];
                b = AB[1];
                AppendArray(ref edge[a],b);
                AppendArray(ref edge[b],a);
            }
            int cnt = 0;
            int neighbor;
            for(int i = 1; i <= n;i++)
            {
                good = true;
                for(int j = 0; j < edge[i].Length; j++)
                {
                    neighbor = edge[i][j];
                    if(Hs[i] <= Hs[neighbor])
                    {
                        good = false;
                    }
                }
                if (good)
                {
                    cnt++;
                }
            }
            Console.WriteLine(cnt);
        }
    }
    //2回位使ったら追加する
    class MyLibrary
    {
        public static int[] ToIntArray(string[] stringArray)
        {
            int length = stringArray.Length;
            int[] intArray = new int[length];
            for (int i = 0; i < length; i++)
            {
                intArray[i] = int.Parse(stringArray[i]);
            }
            return intArray;
        }
        public static void AppendArray(ref int[] intArray, int i)
        {
            int length = intArray.Length;
            Array.Resize(ref intArray, length + 1);
            intArray[length] = i;
        }
    }
}