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 <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <cmath>
#define pi 3.14159265358979323846264338327950

int main() {
	int N, M; std::cin >> N >> M;
	//高さの入力
	std::vector<int> H(N);
	for (auto& r : H) std::cin >> r;
	//道の入力
	std::vector<std::vector<int>> R(2, std::vector<int>(M));
	for (auto i{ 0 }; i < M; ++i)
		std::cin >> R[0][i] >> R[1][i];

	int count{ 0 };
	for (auto num{ 0 }; num < N; ++num) {
		bool flag = false;
		for (int i{ 0 }; i < M; ++i) {
			if (num + 1 == R[0][i])
				if (H[num] <= H[R[1][i] - 1])
					flag = true;
			if (num + 1 == R[1][i])
				if (H[num] <= H[R[0][i] - 1])
					flag = true;
		}
		if (flag == true)
			++count;
	}
	std::cout << N - count << std::endl;

	return 0;
}