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: Wrong Answer)

using System;
using System.Linq;
using System.Collections;
using System.Collections.Generic;
using System.Text;


class Program
{
    static void Main(string[] args)
    {
        // 入力
        //var s = Console.ReadLine();
        //var n = int.Parse(Console.ReadLine());
        long[] nm = Console.ReadLine().Split().Select(long.Parse).ToArray();
        var n = nm[0];
        var m = nm[1];
        long[] h = Console.ReadLine().Split().Select(long.Parse).ToArray();

        long[] a = new long[m];
        long[] b = new long[m];
        for (int i = 0; i < m; i++)
        {
            long[] tmp = Console.ReadLine().Split().Select(long.Parse).ToArray();
            a[i] = tmp[0];
            b[i] = tmp[1];
        }

        var tbd = new Dictionary<long, long>() { };
        for (int i = 0; i < a.Length; i++)
        {
            if (!tbd.ContainsKey(a[i] - 1))
            {
                tbd.Add(a[i] - 1, h[b[i] - 1]);
            }
            else
            {
                if (tbd[a[i] - 1] < h[b[i] - 1])
                {
                    tbd[a[i] - 1] = h[b[i] - 1];
                }
            }

            if (!tbd.ContainsKey(b[i] - 1))
            {
                tbd.Add(b[i] - 1, h[a[i] - 1]);
            }
            else
            {
                if (tbd[b[i] - 1] < h[a[i] - 1])
                {
                    tbd[b[i] - 1] = h[a[i] - 1];
                }
            }
        }

        long ans = 0;
        for (int i = 0; i < tbd.Count; i++)
        {
            if (tbd.ContainsKey(i))
            {
                if (tbd[i] > h[i])
                {
                    ans++;
                }
            }
            else
            {
                ans++;
            }
        }

        // 出力
        Console.WriteLine(ans);
    }
}