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<stdio.h>
#include<string.h>
#include<math.h>
#include <stdlib.h>
#include<ctype.h>

//qsort
int compare_int(const void *a, const void *b)
{
    return *(int*)a - *(int*)b;
}

//比較
int max(int a, int b){
	if (a>b) return a;
	else return b;
}

int min(int a,int b) {
	if(a>b) return b;
	else return a;
}


//main関数
int main() {
	long long N,M;
	scanf("%lld %lld",&N,&M);
	
	long long H[N];
	for(int i=0;i<N;i++) {
		scanf("%lld",&H[i]);
	}
	
	int A[2][N];
	for(int j=0;j<M;j++) {
		scanf("%d %d",&A[0][j],&A[1][j]);
	}
	
	
	int ans[N];
	for(int i=0;i<N;i++) {
		ans[i] = 1;
	}
	
	for(int i=0;i<M;i++){
		if( H[ A[0][i] - 1 ] >= H[ A[1][i] -1] ){
			ans[A[1][i] -1] = 0;
		} 
		if(A[1][i] >= A[0][i] ){
			ans[A[0][i] -1] = 0;
		} 
	}
	int sum = 0;
	for(int i=0;i<N;i++){
		sum += ans[i];
	}
	printf("%d",sum);
}