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>

using namespace std;

#define debug(...) fprintf(stderr, __VA_ARGS__), fflush(stderr)
#define time__(d)for(auto blockTime = make_pair(chrono::high_resolution_clock::now(), true); blockTime.second;debug("%s: %.5lf s\n", d, chrono::duration_cast<chrono::milliseconds>(chrono::high_resolution_clock::now() - blockTime.first).count()/1000.0), blockTime.second = false)

template <typename T>std::string NumberToString ( T Number ){std::ostringstream ss; ss << Number;return ss.str(); } // Usages NumberToString(3)
template <typename T>T StringToNumber ( const string &Text ){istringstream ss(Text);T result;return ss >> result ? result : 0;} // StringToNumber<int>("3")

string StrCpy(string str, long long int start, long long int end){string temp = "";for(long long int i=start; i<=end; i++){temp += str[i];}return temp;}

#define ull unsigned long long int
#define ll long long int 
#define gcd(a, b)  __gcd(a, b)
#define lcm(a, b)  ((a*b)/gcd(a, b))

ll vec[1000009];
ll val[1000009];
ll prio[100009][5];

int solve(){
    ll N,M;
    cin >> N >> M;
    val[0] = 0;
    for(ll i=1; i<=N; i++){
        cin >> val[i];
    }

    ll a,b;

    for(ll i=0; i<M; i++){
        cin >> a >> b;
        if(val[b]>prio[a][0]){
            prio[a][0] = val[b];
        }
        if(val[a]>prio[b][0]){
            prio[b][0] = val[a];
        }
    }

    ll cnt = 0;

    for(ll i=1; i<=N; i++){
        if(prio[i][0] < val[i]){
            //cout << " == " << i  << endl;
            cnt++;
        }
    }

    cout << cnt << endl;

   

    return 0;
}

int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);

    time__("programm time"){
        solve();
	}
    return 0;

}