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 <climits>
#include <cmath>
#include <iostream>
#include <numeric>
#include <queue>
#include <unordered_map>
#include <vector>

const int mod = 1e9 + 7;
const int kmax = 510000;

long long fact[kmax], fact_inv[kmax], inv[kmax];

int gcd(int a, int b)
{
    if (!b) {
        return a;
    }

    return gcd(b, a % b);
}

void init_comb()
{
    fact[0] = fact[1] = 1;
    fact_inv[0] = fact_inv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < kmax; i++) {
        fact[i] = fact[i - 1] * i % mod;
        inv[i] = mod - inv[mod % i] * (mod / i) % mod;
        fact_inv[i] = fact_inv[i - 1] * inv[i] % mod;
    }
}

long long comb(int n, int r)
{
    if (n < r) {
        return 0;
    }

    if (n < 0 || r < 0) {
        return 0;
    }

    return fact[n] * (fact_inv[r] * fact_inv[n - r] % mod) % mod;
}

template <typename T, T N>
class UnionFind {
    T parent_[N];
    T rank_[N];
    T size_[N];

public:
    UnionFind();
    T Root(T idx);
    bool IsSame(T x, T y);
    void Unite(T x, T y);
    T GetSize(T idx);
};

template <typename T, T N>
UnionFind<T, N>::UnionFind()
{
    for (T i = 0; i < N; i++) {
        parent_[i] = i;
        rank_[i] = 0;
        size_[i] = 1;
    }
}

template <typename T, T N>
T UnionFind<T, N>::Root(T idx)
{
    return parent_[idx] == idx ? idx : parent_[idx] = Root(parent_[idx]);
}

template <typename T, T N>
bool UnionFind<T, N>::IsSame(T x, T y)
{
    return Root(x) == Root(y);
}

template <typename T, T N>
void UnionFind<T, N>::Unite(T x, T y)
{
    x = Root(x);
    y = Root(y);
    if (x == y) {
        return;
    }

    if (rank_[x] < rank_[y]) {
        parent_[x] = y;
        size_[y] += size_[x];
    } else {
        parent_[y] = x;
        size_[x] += size_[y];
        if (rank_[x] == rank_[y]) {
            rank_[x]++;
        }
    }
}

template <typename T, T N>
T UnionFind<T, N>::GetSize(T idx)
{
    return size_[Root(idx)];
}

long long pow_mod(long long n, long long p, long long m)
{
    if (p == 0) {
        return 1;
    }

    if (p % 2 == 0) {
        long long t = pow_mod(n, p / 2, m);
        return t * t % m;
    }
    return n * pow_mod(n, p - 1, m) % mod;
}

long long nCr_mod(long long n, long long r)
{
    long long x = 1;
    for (long long i = n - r + 1; i <= n; i++) {
        x *= i;
        x %= mod;
    }

    long long a = 1;
    for (long long i = 1; i <= r; i++) {
        a *= i;
        a %= mod;
    }

    long long y = pow_mod(a, mod - 2, mod) % mod;

    return x * y % mod;
}

int main()
{
    long long n, m;
    std::cin >> n >> m;

    std::vector<long long> h(n);
    for (int i = 0; i < n; i++) {
        std::cin >> h[i];
    }

    std::vector<long long> max(n);
    for (int i = 0; i < m; i++) {
        long long a, b;
        std::cin >> a >> b;
        a--;
        b--;

        max[a] = std::max(max[a], h[b]);
        max[b] = std::max(max[b], h[a]);
    }

    long long count = 0;
    for (int i = 0; i < n; i++) {
        if (h[i] > max[i]) {
            count++;
        }
    }

    std::cout << count << std::endl;
}