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: Runtime Error)

using System;

namespace peaks
{
    class Program
    {
        const int N = 100000;
        static void Main(string[] args)
        {


            int n, m;
            int[] h = new int[N];
            int[]  ma= new int[N];

            string inD0=Console.ReadLine();
            string[] cngD0 = inD0.Split(' ');

            n= int.Parse(cngD0[0]);
            m= int.Parse(cngD0[1]);

            string inD1 = Console.ReadLine();
            string[] cngD1 = inD1.Split(' ');

            for (int i = 0; i <= n-1; i++)                {
                    h[i] = int.Parse(cngD1[i]);
                    ma[i] = 0;
            }
            for (int i = 0; i <= m-1; i++){
                    string inD2 = Console.ReadLine();
                    string[] cngD2 = inD2.Split(' ');

                int a = int.Parse(cngD2[0]);
                    int b= int.Parse(cngD2[1]); 
                    ma[a] = Math.Max(ma[a], h[b]);
                    ma[b] = Math.Max(ma[b], h[a]);
            }
            
            int ans = 0;
            for (int i = 0; i <= n-1; i++)
                {
                    if (h[i] > ma[i]){
                        ans += 1;
                    }
            }
            Console.WriteLine(ans);

        }


    }
}