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<iostream>
#include<vector>
#include<string>
#include<algorithm>
#include<utility>

using ll = long long;
using p = std::pair<ll, ll>;
ll n, m;


int main(){
    std::cin >> n >> m;
    std::vector<p> H(n);
    std::vector<ll> originH(n);
    for(int i = 0; i < n; i++){
        std::cin >> H[i].first;
        originH[i] = H[i].first;
        H[i].second = i;
    }
    
    std::vector<std::vector<ll>> adj(n);
    for(int i = 0; i < m; i++){
        ll a, b;
        std::cin >> a >> b;
        a--; b--;
        adj[a].push_back(b);
        adj[b].push_back(a);
    }

    std::sort(H.begin(), H.end(), std::greater<p>());

    std::vector<bool> isGood(n, true);
    for(int i = 0; i < n; i++){
        ll h = H[i].first;
        ll k = H[i].second;
        if(!isGood[k]) continue;
        bool flag = true;
        for(int j = 0; j < adj[k].size(); j++){
            ll next = adj[k][j];
            ll nextH = originH[next];
            if(h > nextH){
                isGood[next] = false;
            }else{
                flag = false;
            }
        }

        if(!flag){
            isGood[k] = false;
        }
    }

    int cnt = 0;
    for(int i = 0; i < n; i++){
        if(isGood[i]) cnt++;
    }

    std::cout << cnt << std::endl;
    return 0;
}