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: Wrong Answer)

#pragma GCC target("avx")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < ll(n); ++i)
typedef int64_t lint;
using namespace std;
using ll = long long;
using P = pair<ll,ll>;
using lb = long double;
//マクロ
//forループ関係
//引数は、(ループ内変数,動く範囲)か(ループ内変数,始めの数,終わりの数)、のどちらか
//Dがついてないものはループ変数は1ずつインクリメントされ、Dがついてるものはループ変数は1ずつデクリメントされる
#define FOR(i,a,b) for(ll i=a;i<=(ll)(b);i++)
//xにはvectorなどのコンテナ
#define ALL(x) (x).begin(),(x).end() //sortなどの引数を省略したい
#define SI(x) ((ll)(x).size()) //sizeをsize_tからllに直しておく
#define MAX(x) *max_element(ALL(x)) //最大値を求める
#define MIN(x) *min_element(ALL(x)) //最小値を求める
#define debug(x) cerr << #x << ": " << x << "\n"//手動デバッグ
#define debugV(x) for(int i=0;i<((ll)(x).size());i++)cerr << #x <<"at:"<<i<< ": " << x[i] << "\n"//手動デバッグ 
#define YES(c) cout<<((c)?"YES\n":"NO\n"),0
#define Yes(c) cout<<((c)?"Yes\n":"No\n"),0
#define O(p) cout<<p<<"\n",0
#define OS(p) cout<<p<<" ",0
#define ON(p) cout<<p,0
//定数
#define INF 1000000000000 //10^12:極めて大きい値,∞
#define inf 2147483647 //int値の最大値
#define MOD 1000000007 //10^9+7:合同式の法
#define MAXR 100000 //10^5:配列の最大のrange(素数列挙などで使用)
#define endl "\n"//endlを置き換える
//略記
#define PB emplace_back //vectorヘの挿入
#define MP make_pair //pairのコンストラクタ
#define F first //pairの一つ目の要素
#define S second //pairの二つ目の要素
#define CST(x) cout<<fixed<<setprecision(x)//小数点以下の桁数指定

const int dy[4]={1,0,-1,0};
const int dx[4]={0,1,0,-1};//^>v<

template<class T=int> T in(){
    T x;
    cin>>x;
    return (x);
}
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
int gcd(int a, int b) {
    if (b==0) return a;
    else return gcd(b, a%b);
}
int lcm(int a, int b) {
    return a * b / gcd(a, b);
}
int two(int n){
	ll ans=1;
	rep(i,n){
		ans*=2;
	}
	return ans;
}
//------------------------------------------------------------
/*
ll highest(ll n,vector<ll> A,vector<ll> &memo,vector<ll> H){
    ll ans=H[n];
    vector<ll> que(0);
    que.PB(n);
    memo[n]=1;
    while(A[n]!=-1){
        if(memo[A[n]]!=-1)break;
        que.PB(A[n]);
        memo[A[n]]=1;
        ans=max(ans,H[A[n]]);
        n=A[n];
    }
    ll sum=0;
    rep(i,que.size()){
        if(ans==H[que[i]]){
            sum++;
        }
    }
    return sum;
}*/




//------------------------------------------------------------
int main() {
	ios::sync_with_stdio(false);
	std::cin.tie(nullptr);
//------------------------------------------------------------
    int n,m;cin>>n>>m;
    vector<ll> H(n);
    vector<vector<ll>> A(n,vector<ll>(0));
    //vector<ll> memo(n,-1);
    rep(i,n){
        cin>>H[i];
    }
    rep(i,m){
        ll a,b;cin>>a>>b;
        A[a-1].PB(b-1);
        A[b-1].PB(a-1);
    }
    ll ans=0;
    rep(i,n){
        if(A[i].size()!=0){
            bool highest=1;
            rep(j,A[i].size()){
                if(H[i]<=H[A[i][j]]){
                    highest=0;
                    break;
                }
            }
            if(highest)ans++;
        }
    }
    O(ans);
    
    
//------------------------------------------------------------
	return 0;
}