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: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>

int main()
{
	int i, j, ret;
	int N, M;
	int *H;
	int *OK;
	int *A;
	int *B;
	int count;
	int NGhit;
	ret = scanf("%d %d", &N, &M);

	H = (int *) malloc(sizeof(int)*N);
	OK = (int *) malloc(sizeof(int)*N);
	for (i = 0; i < N; i++)
		OK[i] = 1;
	
	for (i = 0; i < N; i++) {
		ret = scanf(" %d", &H[i]);
		// fprintf(stderr, "%d ", H[i]);
		
	}
	// fprintf(stderr, "\n");
	
	A = (int *) malloc(sizeof(int)*M);
	B = (int *) malloc(sizeof(int)*M);
	
	for (i = 0; i < M; i++) {
		ret = scanf(" %d %d\n", &A[i], &B[i]);
		// fprintf(stderr, "%d %d\n", A[i], B[i]);
	}

	for (i = 0; i < N; i++) {
		NGhit = 0;
		/* if i is good */
		for (j = 0; j < M; j++) {
			if (A[j]-1 == i) {
				if (H[B[j]-1] >= H[i]) {
					NGhit = 1;
					break;
				}
			} else if (B[j]-1 == i) {
				if (H[A[j]-1] >= H[i]) {
					NGhit = 1;
					break;
				}
			}
		}
		if (NGhit == 1)
			OK[i] = 0;
	}

	count = 0;
	for (i = 0;i < N; i++) {
		if (OK[i] == 1) {
			count++;
			// fprintf(stderr, "i: %d \n", i);
		}
	}
	printf("%d\n",count);
	fflush(stdout);
	free(H);
	free(OK);
	free(A);
	free(B);
	return EXIT_SUCCESS;
}