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 <string>
#include <vector>

using namespace std;

class Tower {
public: 
  int id;
  vector<Tower *> tonari;
  int h;
  int count;
  Tower() {
    id = 0;
    h = 0;
    count = 0;
  }
};

int main() {
  long long int N, M;
  vector<Tower *> t;

  cin >> N >> M;
  for (int i = 0; i < N; i++)  {
    Tower *tow = new Tower();
    t.push_back(tow);
    int tmp;
    cin >> tmp;
    t[i]->h = tmp;
    t[i]->id = i;
  }
  for (int i = 0; i < M; i++) {
    int a, b;
    cin >> a >> b;
    t[a - 1]->tonari.push_back(t[b - 1]);
    (t[a - 1]->count)++;
    t[b - 1]->tonari.push_back(t[a - 1]);
    (t[b - 1]->count)++;
  }

  int count = 0;
  for (int i = 0; i < N; i++) {
     while (t[i]->count > 0) {
       if (t[i]->tonari[t[i]->count - 1]->h >= t[i]->h) {
         count--;
         break;
       }

       (t[i]->count)--;
     }
     count++;
  }

  cout << count << endl;

  return 0;
}