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 <algorithm>
#include <cmath>
#include <cstdio>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
using namespace std;

#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define REP(i, s, e) for (int i = (int)(s); i < (int)(e); i++)
#define _rep(i, n) for (int i = (int)(n - 1); i >= 0; i--)
#define _REP(i, e, s) for (int i = (int)(e - 1); i >= (int)(s); i--)

#define yes cout << "yes" << endl;
#define Yes cout << "Yes" << endl;
#define YES cout << "YES" << endl;

#define no cout << "no" << endl;
#define No cout << "No" << endl;
#define NO cout << "NO" << endl;

#define ll long long
#define ull unsigned long long
#define vi vector<int>

const double PI = 3.14159265358979323846;

ll gcd(ll a, ll b) {
    if (a < b) return gcd(b, a);
    ll r;
    while ((r = a % b)) {
        a = b;
        b = r;
    }
    return b;
}

int main() {
    long long N, M, A, B, ans = 0;
    cin >> N >> M;
    vector<long long> H(N), H2(N);
    rep(i, N) cin >> H[i];
    rep(i, M) {
        cin >> A >> B;
        H2[A - 1] = max(H2[A - 1], H[B - 1]);
        H2[B - 1] = max(H2[B - 1], H[A - 1]);
    }

    for (int i = 0; i < N; i++) {
        ans += H[i] > H2[i];
    }
    cout << ans << endl;
}