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<stdio.h>
int main(){
	int Tai_num, Road_num;
	scanf("%d", &Tai_num);
	scanf("%d", &Road_num);
	int height[5000];
	int road[5000][2];
	int delete[500];
	int i = { 0 };


	//高度和道路数组
	for (int i = 0;i < Tai_num;i++)
		scanf("%d", &height[i]);

	for (int i = 0;i < Road_num;i++)
		for (int j = 0;j < 2;j++)
			scanf("%d", &road[i][j]);

	for (int i = 0;i < Road_num;i++)
	{
		int first = road[i][0];
		int second = road[i][1];

		if (height[first] > height[second])
			delete[i++] = second;
		else
			delete[i++] = first;        //获得对应不满足要求的下标
	}
	
	for (int i = 0;i < Tai_num;i++)
	{
		for (int j = 0;j < 5000;j++)
			if (i == delete[j]) {
				Tai_num--;
				break;
			}
	}
	printf("%d", Tai_num);

}