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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
#ifdef ENABLE_DEBUG
#define dump(a) cerr<<#a<<"="<<a<<endl
#define dumparr(a,n) cerr<<#a<<"["<<n<<"]="<<a[n]<<endl
#else
#define dump(a) 
#define dumparr(a,n) 
#endif
#define FOR(i, a, b) for(ll i = (ll)a;i < (ll)b;i++)
#define For(i, a) FOR(i, 0, a)
#define REV(i, a, b) for(ll i = (ll)b-1LL;i >= (ll)a;i--)
#define Rev(i, a) REV(i, 0, a)
#define REP(a) For(i, a)
#define SIGN(a) (a==0?0:(a>0?1:-1))

typedef long long int ll;
typedef unsigned long long ull;
typedef unsigned int uint;
typedef pair<ll, ll> pll;
typedef pair<ll,pll> ppll;
typedef vector<ll> vll;
typedef long double ld;
typedef pair<ld,ld> pdd;

const ll INF=(1LL<<50);
#if __cplusplus<201700L
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;
}
#endif
template<class T>
bool chmax(T& a,const T& b){
  if(a<b){
    a=b;
    return true;
  }
  return false;
}
template<class T>
bool chmin(T& a,const T& b){
  if(a>b){
    a=b;
    return true;
  }
  return false;
}
template<class S,class T>
std::ostream& operator<<(std::ostream& os,pair<S,T> a){
  os << "(" << a.first << "," << a.second << ")";
  return os;
}
template<class T>
std::ostream& operator<<(std::ostream& os,vector<T> a){
  os << "[ ";
  REP(a.size()){
    os<< a[i] << " ";
  }
  os<< " ]";
  return os;
}


void solve(long long N, long long M, std::vector<long long> H, std::vector<long long> A, std::vector<long long> B){
  set<ll> dai;
  REP(N){
    dai.insert(i);
  }
  REP(M){
    if(H[A[i]]>H[B[i]]){
      dai.erase(B[i]);
    }else if(H[A[i]]<H[B[i]]){
      dai.erase(A[i]);
    }else{
      dai.erase(A[i]);
      dai.erase(B[i]);
    }
  }
  cout<<dai.size()<<endl;
}

int main(){
  cout<<setprecision(1000);
    long long N;
    scanf("%lld",&N);
    long long M;
    scanf("%lld",&M);
    std::vector<long long> H(N);
    for(int i = 0 ; i < N ; i++){
        scanf("%lld",&H[i]);
    }
    std::vector<long long> A(M);
    std::vector<long long> B(M);
    for(int i = 0 ; i < M ; i++){
        scanf("%lld",&A[i]);
        scanf("%lld",&B[i]);
        --A[i];--B[i];
    }
    solve(N, M, std::move(H), std::move(A), std::move(B));
    return 0;
}