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: Runtime Error)

#include <iostream>
#include <string>
#include <vector>

#define rep(i,n) for(int i=0; i <(n); i++)

using namespace std;

struct edge{
    int to;
    int cost;
};

vector<edge> graph[100000];

int main(void){

    int N, M;
    cin >> N >> M;

    static long int H[100000];
    rep(i,N) cin >> H[i+1];

    for(int i=0; i < M; i++){
        int A,B;
        cin >> A >> B;

        int cost1 = H[A] - H[B];
        edge e1 = {B, cost1};
        graph[A].push_back(e1);

        int cost2 = H[B] - H[A];
        edge e2 = {A, cost2};
        graph[B].push_back(e2);
    }

    int count = 0;
    for(int i=1; i < N; i++){
        int is_peak = true;

        for(int j=0; j < graph[i].size(); j++){
            edge e = graph[i][j];
            if(e.cost < 0){
                is_peak = false;
            }
        }
        if(is_peak == true) count++;
    }

    cout << count;

    return 0;
}