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++ (Clang 10.0.0) (judge verdict: Accepted)

#include <cstdio>
#include <vector>
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int MAXN = 100010;
inline int read() {
    int n = 0, f = 1; char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-') f = -f;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        n = 10 * n + (c ^ '0');
        c = getchar();
    }
    return f * n;
}
int val[MAXN];
vector<int> edge[MAXN];
int main() {
    int n = read(), m = read();
    for (int i = 1; i <= n; i++) val[i] = read();
    for (int i = 1; i <= m; i++) {
        int u = read(), v = read();
        edge[u].push_back(v);
        edge[v].push_back(u);
    }
    int ans = 0;
    for (int u = 1; u <= n; u++) {
        bool ok = 1;
        for (int i = 0; i < edge[u].size(); i++) {
            int v = edge[u][i];
            if (val[v] >= val[u]) {
                ok = 0;
            }
        }
        ans += ok;
    }
    printf("%d\n", ans);
    return 0;
}