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 <iomanip>
#include <string>
#include <vector>
#include <queue>
#include <unordered_map>
#include <set>
#include <algorithm>
#include <utility>
#include <cmath>
using namespace std;

typedef unsigned int uint;
typedef long long llong;
typedef unsigned long long ullong;
// -2.1e-9 <= int <= 2.1e9
// -9.2e-18 <= long long <= 9.2e18
// 0 <= unsigned int <= 4.2e9
// 0 <= unsigned long long <= 18.4e18
// cout << fixed << setprecision(10);

int main()
{
    uint N, M;
    cin >> N >> M;
    vector<int> H(N+1, 0);
    vector<vector<int>> G(N+1);
    for (size_t i=1; i<H.size(); i++)
        cin >> H[i];

    for (size_t r=0; r<M; r++) {
        uint i;
        uint j;
        cin >> i >> j;
        G[i].push_back(j);
        G[j].push_back(i);
    }
    int good = 0;
    for (size_t i=1; i<G.size(); i++) {
        if (G[i].size() == 0) {
            ++good;
            continue;
        }
        else {
            bool judge = true;
            for (size_t j=0; j<G[i].size(); j++) {
                if (H[i] <= H[G[i][j]]) {
                    judge = false;
                    break;
                }
            }
            if (judge) ++good;
        }
    }
    cout << good << endl;
}