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++ (Clang 10.0.0) (judge verdict: Accepted)

#include<iostream>
#include<cstdio>
#define MAX 100000
using namespace std;
int main(){
  int N, M, count, Root[MAX], A, B;
  long long  H[MAX];
  count = 0;
  scanf("%d %d", &N, &M);
  for(int i = 0; i < N; i++){
    scanf("%lld", &H[i]);
    Root[i] = 1;
  }
 
  for(int i = 0; i < M; i++){
    scanf("%d %d", &A, &B);
    if(H[A - 1] < H[B - 1]) Root[A - 1] = 0;
    else if( H[B - 1] < H[A - 1]) Root[B - 1] = 0;
    else if( H[A - 1] == H[B - 1]) Root[A - 1] = Root[B - 1] = 0;
    
  }
  
  for(int i = 0; i < N; i++){
    if(Root[i] == 1) count++;
  }
  printf("%d\n", count);
  return 0;
}