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: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;
#define ll long long
#define FOR(i, a, b) for (ll i = (a); i < (b); ++i)
#define REP(i, n) FOR(i, 0, n)
#define PRINT(item) cout << (item) << endl;
#define PRINT_VEC(V)                                                           \
    for (auto v : (V))                                                         \
        cout << v << ' ';
const int MOD = 1000000007;
/* cout << fixed << setprecision(10) << decimal << endl; */

int main() {
    /* code */
    ll N, M;
    cin >> N >> M;
    vector<ll> H(N);
    REP(i, N) { cin >> H[i]; }
    using AB = pair<ll, ll>;

    vector<bool> Ans(N, true);
    REP(i, M) {
        int A, B;
        cin >> A >> B;
        A--;
        B--;
        //
        if (H[A] > H[B]) {
            // firstのほうが高い → secondはだめは展望台
            Ans[B] = false;
        }
        if (H[A] < H[B]) {
            Ans[A] = false;
        }
    }
    int res = 0;
    REP(i, N) {
        if (Ans[i])
            res++;
    }
    cout << res << endl;

    return 0;
}