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 <cstring>
#include <cmath>
#include <vector>
#include <algorithm>
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <set>
#include <map>
#include <iterator>
#include <stack>
#include <string.h>
#include <cstdlib>
#include <queue>
#include <list>
#include <string>

using namespace std;

bool ans[100005] = {0};

int main()
{
    int n,m;
    long long h[100005];
    
    cin >> n >> m;
    for (int i = 0;i < n;i++){
        cin >> h[i];
    }
    
    for (int i = 0;i < m;i++){
        int a,b;
        cin >> a >> b;
        a--;
        b--;
        if (h[a] > h[b]){
            ans[b] = true;
        }
        else if (h[a] == h[b]){
            ans[a] = true;
            ans[b] = true;
        }
        else {
            ans[a] = true;
        }
    }
    
    int t = 0;
    for (int i = 0;i < n;i++){
        if (ans[i]){
            t++;
        }
    }
    
    cout << n-t << endl;
}