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 rep(i,n) for(int i=0; i<(int)(n); i++)
using ll = long long;
using P = pair<int, int>;

int main(){
    int N, M;
    cin >> N >> M;
    vector<int> H(N), A(M), B(M);
    rep(i, N) cin >> H.at(i);
    rep(i, M) cin >> A.at(i) >> B.at(i);

    vector<int> cnt(N);
    rep(i, M){
        if (H.at(A.at(i) - 1) > H.at(B.at(i) - 1)){
            cnt.at(B.at(i) - 1)++;
        }
        if (H.at(A.at(i) - 1) < H.at(B.at(i) - 1)){
            cnt.at(A.at(i) - 1)++;
        }
        if (H.at(A.at(i) - 1) == H.at(B.at(i) - 1)){
            ++cnt.at(A.at(i) - 1);
            ++cnt.at(B.at(i) - 1);
        }
        
    }
    int ans = 0;
    rep(i, N){
        if (cnt.at(i) == 0) ++ans;
    }
    
    cout << ans << endl;
    return 0;
}