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>
#define REP(i, n) for(int i = 0; i < (n); i++)
#define REP1(i, n) for(int i = 1; i <= (n); i++)
#define REPD(i,a,b) for (int i=(a);i<(b);i++)
#define ALL(v) (v).begin(), (v).end()
using namespace std;
typedef long long ll;
typedef vector<string> vs;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef set<int> si;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef pair<int, int> pi;
typedef queue<pi> qpi;
typedef pair<ll, ll> pll;
typedef queue<pll> qpll;
typedef vector<pi> vpi;
typedef vector<pll> vpll;
const int mod = 1000000007;
const int INF = 1001001001;
// 小数点　<< fixed << setprecision(0) <<
// sort降順　sort(ALL(),greater<int>());
// 円周率　M_PI
// 文字判定　isupper islower
// 順列　do {} while(next_permutation(ALL(X)));
// 最大値　LLONG_MAX
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
struct edge {
    int from; //出発点
    int to;   //到達点
    int cost; //移動コスト
};
typedef struct edge se;
typedef vector<edge> ve;
unsigned euclidean_gcd(unsigned a, unsigned b) {
  if(a < b) return euclidean_gcd(b, a);
  unsigned r;
  while ((r=a%b)) {
    a = b;
    b = r;
  }
  return b;
}
int main() {
    ll n,m;
    cin >> n >> m;
    vll h(n);
    REP(i,n) cin >> h[i];
    ll a,b;
    vector<bool> check(n,true);
    REP(i,m) {
        cin >> a >> b;
        a--;
        b--;
        if(h[a]<h[b]) {
            check[a]=false;
        } else if(h[a]>h[b]) {
            check[b]=false;
        } else {
            check[a]=false;
            check[b]=false;
        }
    }
    ll ans=0;
    REP(i,n) {
        if(check[i]) ans++;
    }
    cout << ans << endl;
  return 0;
}