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: Wrong Answer)

/*#include <bits/stdc++.h>
using namespace std;
int main(){
  
  int obs,rd;
  cin >> obs >> rd;
 // vector<long long> hgt(obs);
  map<int,long long> o_h;
  map<int,int> res;
  map<int,bool> mp;
  int a,b;
  
  for(int i = 0; i < obs; i++){
    long long x;
    cin >> x;
  //  hgt[i] = x;
    o_h[i+1] = x;
  }
 // cout << res[1];
  
  for(int i = 0; i < rd; i++){
    cin >> a >> b;
    if(o_h[a] < o_h[b]){
      res[a] = -1;
    }
    if(o_h[b] < o_h[a])res[b] = -1;
    mp[a] = mp[b] = true;
  }
  int cnt = 0;
  for(int i = 1; i <= obs; i++){
    if(res[i] != -1 && mp[i] == true)cnt++;
  }
  cout << cnt;
  
}*/
#include <bits/stdc++.h>
using namespace std;
int main(){
  
  int obs,rd;
  cin >> obs >> rd;
 // vector<long long> hgt(obs);
  map<int,long long> o_h;
  map<int,int> res;
  map<int,bool> mp;
  int a,b;
  
  for(int i = 0; i < obs; i++){
    long long x;
    cin >> x;
  //  hgt[i] = x;
    o_h[i+1] = x;
  }
 // cout << res[1];
  
  for(int i = 0; i < rd; i++){
    cin >> a >> b;
    if(o_h[a] < o_h[b]){
      res[a] = -1;
    }
    if(o_h[b] == o_h[a])res[b] = -1;
  //  mp[a] = mp[b] = true;
  }
  int cnt = 0;
  for(int i = 1; i <= obs; i++){
    if(res[i] != -1)cnt++;
  }
  cout << cnt;
  
}