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)

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

template<typename T>
istream& operator >> (istream& istr, vector<T>& v){
    for(T& x: v) istr >> x;
    return istr;
}

template<typename T, typename U>
istream& operator >> (istream& istr, pair<T,U>& p){
    istr >> p.first >> p.second;
    return istr;
}

template<typename T>
ostream& operator << (ostream& ostr, vector<T>& v){
    if(!v.empty()){
        ostr << v.front();
        for(auto itr = ++v.begin(); itr != v.end(); itr++)
            // ostr << *itr;
            ostr << " " << *itr;
    }
    return ostr;
}

template<typename T, typename U>
ostream& operator << (ostream& ostr, pair<T,U>& p){
    // ostr << p.first << p.second;
    // ostr << p.first << ", " << p.second;
    ostr << "(" << p.first << ", " << p.second << ")";
    return ostr;
}

template<typename T>
ostream& operator << (ostream& ostr, vector<vector<T>>& vv){
    if(!vv.empty()){
        ostr << vv.front();
        for(auto itr = ++vv.begin(); itr != vv.end(); itr++)
            ostr << endl << *itr;
    }
    return ostr;
}

template<typename T, typename U>
ostream& operator << (ostream& ostr, vector<pair<T,U>>& vp){
    if(!vp.empty()){
        ostr << vp.front();
        for(auto itr = ++vp.begin(); itr != vp.end(); itr++)
            ostr << endl << *itr;
    }
    return ostr;
}

typedef long long ll;
typedef vector<ll> vll;
typedef vector<char> vc;
typedef vector<double> vd;
// vll v(n); v[i]=x;
typedef vector<vll> vvll;
typedef vector<vc> vvc;
typedef vector<vd> vvd;
// vvll v(m,vll(n)); v[i][j]=x;
typedef pair<ll,ll> pll;
typedef pair<double,double> pd;
typedef vector<pll> vpll;
typedef vector<pd> vpd;
// vpll v(n); v[i]=make_pair(x,y); v[i].first/second;
// v.front() : v[0]
// v.back()  : v[v.size()-1]

#define PI 3.1415926535897932
#define EPS 1e-12
#define INF ((ll)1e+12)
#define REP(i,n) for(ll i=0; i<(n); i++)
#define RREP(i,n) for(ll i=(n)-1; i>=0; i--)
#define FOR(i,a,b) for(ll i=(a); i<=(b); i++)
#define FORR(i,a,b) for(ll i=(b); i>=(a); i--)
#define fix(x) cout << fixed << setprecision(x)
#define dump(x) cout << #x << " = " << (x) << endl
#define all(x) (x).begin(),(x).end()

template<typename T> T sum(vector<T>& v){return accumulate(all(v), (T)0);}
template<typename T> inline void sort(vector<T>& v){sort(all(v));}
template<typename T> inline void rsort(vector<T>& v){sort(all(v), greater<T>());}
template<typename T> inline void chmin(T& a, T b){if(a>b) a=b;}
template<typename T> inline void chmax(T& a, T b){if(a<b) a=b;}

int main(){
    ll n, m, ans = 0;
    cin >> n >> m;
    vll h(n);
    vvll r(n);

    cin >> h;
    REP(i,m){
        ll a, b;
        cin >> a >> b;
        a--;
        b--;
        r[a].emplace_back(b);
        r[b].emplace_back(a);
    }
    REP(i,n){
        REP(j, r[i].size()){
            if(h[r[i][j]] >= h[i]) break;
            if(j == r[i].size()-1) ans++;
        }
    }
    cout << ans << endl;
    return 0;
}