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: Time Limit Exceeded)

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

class Program
{
    public static void Main( )
    {
        var input = Console.ReadLine( ).Split( ).Select( int.Parse ).ToArray( );
        int N = input[ 0 ];
        int M = input[ 1 ];
        var H = Console.ReadLine( ).Split( ).Select( int.Parse ).ToArray( );

        var graph = Enumerable.Range( 1, N ).Select( x => new int[ N ] ).ToArray( );    // int[M][2]
        for ( int i = 0; i < M; i++ )
        {
            var e = Console.ReadLine( ).Split( ).Select( int.Parse ).ToArray( );
            graph[ --e[ 0 ] ][ --e[ 1 ] ] = graph[ e[ 1 ] ][ e[ 0 ] ] = 1;      // 0 origin調整
        }

        var result = new bool[ N ];
        for ( int i = 0; i < N; i++ )
        {
            var row = graph[ i ];
            var reachable = row.Select( ( val, idx ) => (val, idx) ).Where( tap => tap.val == 1 ).Select( tap => H[ tap.idx ] ).ToArray( );
            bool isGood = reachable.All( x => H[ i ] > x ) || reachable.Length == 0;
            result[ i ] = isGood;
        }

        Console.WriteLine( result.Where( x => x ).Count( ) );
    }
}