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 <bits/stdc++.h>
using namespace std;

#define each(i, c) for (auto& i : c)
#define mkp(a, b) make_pair(a, b)

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> Pll;
const ll MOD = 1e9+7;

template<typename P, typename Q> ostream& operator << (ostream& os, pair<P, Q> p) { os << "(" << p.first << ": " << p.second << ")"; return os; }
template<typename T> ostream& operator << (ostream& os, vector<T> v) { os << "("; each (i, v) os << i << ", "; os << ")"; return os; }
template<typename K, typename V> ostream& operator << (ostream& os, map<K, V> m) { os << "{"; each (i, m) os << i << ", "; os << "}"; return os; }

int main() {
  ll n, m;
  cin >> n >> m;

  vector<ll> h(n);
  for (ll i = 0; i < n; ++i) {
    cin >> h[i];
  }

  vector<vector<ll>> g(n);
  for (ll i = 0; i < m; ++i) {
    ll a, b;
    cin >> a >> b;
    a--; b--;
    g[a].push_back(h[b]);
    g[b].push_back(h[a]);
  }

  ll ans = 0;
  for (ll i = 0; i < n; ++i) {
    if (g[i].empty()) {
      ++ans;
      continue;
    }

    sort(g[i].begin(), g[i].end(), greater<ll>());
    if (h[i] > g[i][0]) ++ans;
  }

  cout << ans << endl;

  return 0;
}