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;
using System.Text;

public class MainClass
{
	public const long Giri = 1000000007;
	public const long DpInf = 99999999999999;
	
	public class View
	{
		public int Index {get;set;}
		public long Height {get;set;}
		public List<int> Roots {get;set;}
	}
	public static void Main(string[] args)
	{
		var p = Input().ToInt();
		var n = p[0];
		var m = p[1];
		var q = Input();
		var h = new long[n];
		var v = new View[n];
		for(int i = 0; i < n; i++)
		{
			h[i] = q[i].ToLong();
			v[i] = new View(){Index=i, Height = h[i], Roots = new List<int>()};
		}
		for(int i = 0; i < m; i++)
		{
			p = Input().ToInt(-1);
			v[p[0]].Roots.Add(p[1]);
			v[p[1]].Roots.Add(p[0]);
			//u.Unite(p[0], p[1]);
		}
		long wa = 0;
		for(int i = 0; i < n; i++)
		{
			long maxHeight = 0;
			foreach(var k in v[i].Roots)
			{
				maxHeight = Math.Max(maxHeight, v[k].Height);
			}
			if(maxHeight < v[i].Height)
				wa++;
		}
		Console.WriteLine(wa);
		
	}
	
	
	static long GCD(long a, long b)
	{
		var v = new[] { a, b };
		while (v[1] != 0) { v = new[] { v[1], v[0] % v[1] }; }
		return v[0];
	}
	public static string[] Input() { return Console.ReadLine().Split(); }
}

public static class Smarter
{
	public static int ToInt(this string str) { return int.Parse(str); }
	public static long ToLong(this string str) { return long.Parse(str); }
	public static int[] ToInt(this string[] str, int offset = 0)
	{
		var r = new int[str.Length];
		for(int i = 0; i < str.Length; i++) r[i] = str[i].ToInt() + offset;
		return r;
	}
}