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.IO.Pipes;
using System.Linq;
using System.Net;
using System.Numerics;
using System.Text.RegularExpressions;
using System.Threading;

namespace Atcoder
{
	class Program
	{
		//string s = Console.ReadLine();
		//int[] intArray = Console.ReadLine().Split(' ').Select(i => int.Parse(i)).ToArray();
		//int H = intArray[0];
		//int W = intArray[1];
		//int N = int.Parse(Console.ReadLine());
		//List<int> A = new List<int>(Console.ReadLine().Split(' ').Select(i => int.Parse(i)).ToArray());
		//Boolean ok = false;

		//Console.WriteLine(ans);
		//Console.WriteLine("Yes");
		//Console.WriteLine("No");
		//Console.ReadKey();
		//Console.WriteLine(cnt);

		//string[] strArray = Console.ReadLine().Split(' ').ToArray();
		//string S = strArray[0];
		//string T = strArray[1];
		static void Main(string[] args)
		{

			int[] NM = Console.ReadLine().Split(' ').Select(i => int.Parse(i)).ToArray();
			int N = NM[0];
			int M = NM[1];
			int[] H = Console.ReadLine().Split(' ').Select(i => int.Parse(i)).ToArray();
			int mm = M * 2;

			int[,] AB = new int[mm, 2];
			
			for (int i = 0;i < M; i++)
            {
				int[] tmp = Console.ReadLine().Split(' ').Select(i => int.Parse(i)).ToArray();
				AB[i, 0] = tmp[0];
				AB[i, 1] = tmp[1];
				AB[mm - 1 - i, 0] = tmp[1];
				AB[mm - 1 - i, 1] = tmp[0];
			}

			bool[] ans = new bool[N];
			for (int i = 0; i < N; i++)
			{
				ans[i] = true;
			}

			for (int i = 0; i < mm; i++)
			{
				if (H[AB[i, 0] - 1] <= H[AB[i, 1] - 1])
				{
					ans[AB[i, 0] - 1] = false;
				}
			}

			int cnt = ans.Count(x => x == true);
			Console.WriteLine(cnt);
		}

		/// <summary>
		/// 回文
		/// </summary>
		/// <param name="Target"></param>
		/// <param name="len"></param>
		/// <returns></returns>
		static bool Palindrome(string Target , int len)
        {
			bool yes = true;

			for(int i = 0;i < len; i++)
            {
				if(Target[i] != Target[len - 1 - i])
                {
					yes = false;
					break;
                }
            }
			return yes;
        }

		/// <summary>
		/// アナグラム
		/// </summary>
		/// <param name="word1"></param>
		/// <param name="word2"></param>
		/// <returns></returns>
		static bool IsAnagramOf(string word1, string word2)
		{
			return word1.OrderBy(x => x).SequenceEqual(word2.OrderBy(x => x));
		}
		/// <summary>
		/// 最小公倍数
		/// </summary>
		/// <param name="a"></param>
		/// <param name="b"></param>
		/// <returns></returns>
		public static BigInteger Lcm(BigInteger a, BigInteger b)
		{
			return a * b / Gcd(a, b);
		}

		/// <summary>
		/// 最大公約数
		/// </summary>
		/// <param name="a"></param>
		/// <param name="b"></param>
		/// <returns></returns>
		public static BigInteger Gcd(BigInteger a, BigInteger b)
		{
			if (a < b)
				// 引数を入替えて自分を呼び出す
				return Gcd(b, a);
			while (b != 0)
			{
				var remainder = a % b;
				a = b;
				b = remainder;
			}
			return a;
		}
	}
}