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 <vector>
#include <algorithm>
#include <iomanip>
#include <string>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <tuple>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <cassert>
#include <cstdint>
#include <cctype>
#include <numeric>
#include <bitset>
#include <functional>

using namespace std;

using ll =  long long;
using Pll = pair<ll, ll>;
using Pii = pair<int, int>;

constexpr int INF = 1 << 30;
constexpr ll LINF = 1LL << 60;
constexpr ll MOD = 1000000007;
constexpr long double EPS = 1e-10;
constexpr int dyx[4][2] = {
    { 0, 1}, {-1, 0}, {0,-1}, {1, 0}
};

int main() {
    ios::sync_with_stdio(false); cin.tie(nullptr);
    int n, m, a, b;
    cin >> n >> m;
    vector<int> h(n);
    for(int i=0;i<n;++i) cin >> h[i];

    vector<bool> is_good(n, true);
    for(int i=0;i<m;++i) {
        cin >> a >> b;
        --a; --b;
        if(h[a] > h[b]) {
            is_good[b] = false;
        } else if(h[a] < h[b]) {
            is_good[a] = false;
        } else {
            is_good[a] = false;
            is_good[b] = false;
        }
    }

    int ans = 0;
    for(int i=0;i<n;++i) if(is_good[i]) ++ans;
    cout << ans << endl;
}