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;

#define endl "\n"
#define FSP(n) cout << fixed << setprecision(n);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
static inline ll gcd(ll a, ll b) { return a % b == 0 ? b : gcd(b, a % b);}
static inline ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
#define prepare cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);

signed main() {
    prepare
    int N, M;
    cin >> N >> M;
    vector<int> H(N, 0);
    for (int i = 0; i < N; ++i) {
        cin >> H[i];
    }
    vector<int> A(M, 0);
    vector<int> B(M, 0);
    for (int j = 0; j < M; ++j) {
        cin >> A[j] >> B[j];
    }
    vector<bool> T(N, true);
    vector<int> cnt(N, 1);
    for (int k = 0; k < M; ++k) {
        int X = A[k] - 1;
        int Y = B[k] - 1;

        if (H[X] < H[Y]) {
            T[X] = false;
        }
        else if (H[Y] < H[X]) {
            T[Y] = false;
        }
        else if (H[X] == H[Y]) {
            T[X] = false;
            T[Y] = false;
        }

    }
    int ans = 0;
    for (int l = 0; l < N; ++l) {
        if (T[l] == true) ans++;
    }
    cout << ans << endl;
}