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 <bits/stdc++.h>
using namespace std;

int main() {
  int n, m;
  cin >> n >> m;
  vector<int> h(n);
  vector<vector<int>> judge(n);
  for(int i = 0; i < n; i++) {
    int a;
    cin >> a;
    h.at(i) = a;
  }
  
  for(int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    judge.at(a).push_back(b);
    judge.at(b).push_back(a);
  }
  
  /*for(int i = 0; i < n; i++) {
    cout << judge.at(i).size() << endl;
  }*/
  
  int ans = 0;
  for(int i = 0; i < n; i++) {
    if(judge.at(i).size() == 0) {
      ans++;
      continue;
    }
    for(int j = 0; j < judge.at(i).size(); j++) {
      if(h.at(i) <= h.at(judge.at(i).at(j))) {
        break;
      }
      if(j == judge.at(i).size()-1 && h.at(i) > h.at(judge.at(i).at(j))) {
        ans++;
      }
    }
  }
  
  cout << ans << endl;
  
  return 0;
}