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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <vector>

int main()
{
  unsigned int N, M;
  std::cin >> N >> M;
  std::vector<unsigned long long> highArray(N);
  std::vector<bool> goodArray(N, true);
  for(auto& high : highArray)
  {
    unsigned long long H;
    std::cin >> H;
    high = H;
  }
  
  for(int i = 0; i < M; ++i)
  {
    unsigned int A, B;
    std::cin >> A >> B;
    
    const int aIdx = A - 1;
    const int bIdx = B - 1;
    if(goodArray[aIdx])
	    goodArray[aIdx] = highArray[aIdx] > highArray[bIdx];
    
    if(goodArray[bIdx])
	    goodArray[bIdx] = highArray[aIdx] < highArray[bIdx];
  }
  
  int cnt = 0;
  for(const auto& good : goodArray)
  {
    if(good)
      ++cnt;
  }
  
  std::cout << cnt << std::endl;
  return 0;
}