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 < (int)(n); i++)
#define REPR(i, n) for(int i = (int)(n); i >= 0; i--)
#define FOR(i, s, n) for (int i = (s); i < (int)(n); i++)
#define ALL(v) v.begin(), v.end()
using namespace std;
typedef long long ll;

ll N, K, T, W, H, M, C;

const int MAX = 5100000;
const ll MOD = 1e9 + 7;

const double PI=3.14159265358979323846;


long long fac[MAX], finv[MAX], inv[MAX];

vector<double> cx, cy, c;
 ll MM = 1e12;

void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

ll modinv(ll a) {
    ll b = MOD, u = 1, v = 0;
    while (b) {
        ll t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= MOD;
    if (u < 0) u += MOD;
    return u;
}

ll modpow(ll x, ll n) {
  ll ret = 1;
  while(n > 0) {
    if(n & 1) (ret *= x) %= MOD;
    (x *= x) %= MOD;
    n >>= 1;
  }
  return ret;
}

struct Eratos {
    vector<int> primes;
    vector<bool> isprime;
    vector<int> min_factor;

    Eratos(int MAX) : primes(),
                      isprime(MAX+1, true),
                      min_factor(MAX+1, -1) {
        isprime[0] = isprime[1] = false;
        min_factor[0] = 0, min_factor[1] = 1;
        for (int i = 2; i <= MAX; ++i) {
            if (!isprime[i]) continue;
            primes.push_back(i);
            min_factor[i] = i;
            for (int j = i*2; j <= MAX; j += i) {
                isprime[j] = false;
                if (min_factor[j] == -1) min_factor[j] = i;
            }
        }
    }

    // prime factorization
    vector<pair<int,int>> prime_factorize(int n) {
        vector<pair<int,int> > res;
        while (n != 1) {
            int prime = min_factor[n];
            int exp = 0;
            while (min_factor[n] == prime) {
                ++exp;
                n /= prime;
            }
            res.push_back(make_pair(prime, exp));
        }
        return res;
    }
};

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;
}

ll lcm(ll a, ll b)
{
   return a * b / gcd(a, b);
}

  

struct UnionFind {
    vector<int> par; 

    UnionFind(int N) : par(N) { 
        for(int i = 0; i < N; i++) par[i] = i;
    }

    int root(int x) { 
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }

    void unite(int x, int y) {
        int rx = root(x);
        int ry = root(y); 
        if (rx == ry) return; 
        par[rx] = ry; 
    }

    bool same(int x, int y) { 
        int rx = root(x);
        int ry = root(y);
        return rx == ry;
    }
};

int main(){
    cin >>N>> M;
    vector<ll> h(N);
    vector<vector<ll>> g(N, vector<ll> (0));
    REP(i,N) cin >> h[i];
    REP(i,M){
        int a,b;
        cin >> a >>b;
        g[a-1].push_back(h[b-1]);
        g[b-1].push_back(h[a-1]);
    }
    int ans = 0;
    for(int i = 0; i < N; ++i){
        ans ++;
        if(g[i].size() == 0) continue;
            sort(ALL(g[i]), greater<ll>());
            if(h[i] <= g[i][0]) ans--;
    }
    cout << ans;
    return 0;
}