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;
public class Hello
{
    public static void Main()
    {
        
        var NM = Console.ReadLine().Split().Select(e => int.Parse(e)).ToArray();
        //展望台の数
        var N = NM[0];
        //道の数
        var M = NM[1];
        //高さを高さ配列に保持する
        var height = Console.ReadLine().Split().Select(e => int.Parse(e)).ToArray();
        var conn = new Dictionary<int,List<int>>();
        //pathをもたない展望台を記録するための配列
        var havePath = new bool[N];
        for(int num = 0;num < M;num++)
        {
            //pathの入力
            var path = Console.ReadLine().Split().Select(e => int.Parse(e)).ToArray();
            var A = path[0];
            var B = path[1];
            //起点となる展望台がすでに存在すれば行き先を追加
            // 存在しなければ生成する
            if(!conn.ContainsKey(A))
            {
                conn[A] = new List<int>();
                conn[A].Add(B);
            }
            else
            {
                conn[A].Add(B);
            }
            if(!conn.ContainsKey(B))
            {
                conn[B] = new List<int>();
                conn[B].Add(A);
            }
            else
            {
                conn[B].Add(A);
            }
            //行き先をもつ展望台を記録する
            havePath[A-1]=true;
            havePath[B-1]=true;
        }
        var ans = 0;
        //生成したDictionaryをすべてループする
        foreach(var c in conn)
        {
            //Keyとなる展望台の標高を保持する
            var heightNow = height[c.Key-1];
            var flg = true;
            foreach(var val in c.Value)
            {
                //標高が他の展望台以下であるなら回答になりえない
                if(heightNow <= height[val-1])
                {
                    flg =false;
                    break;
                }
            }
            if(flg)
            {
                ans++;
            }
        }
        Console.WriteLine(ans + havePath.Count(e => e == false));
    }
}