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: Accepted)

#include<iostream>
#include<set>
#include<vector>
#include<cmath>
#include<list>
#include<algorithm>

#define ll long long

using namespace std;

int main() {
	ll n, m;
	vector<ll> h,a,b;
	ll temp,t1,t2;
	vector<int> q(100000,1);
	ll cnt = 0;
	cin >> n >> m;
	for (ll nn = 0; nn < n;++nn) {
		cin >> temp;
		h.push_back(temp);
	}
	for (ll mm = 0; mm < m; ++mm) {
		cin >> t1 >> t2;
		//a.push_back(t1);
		//b.push_back(t2);
		if (h[t1 - 1] < h[t2 - 1]) q[t1 - 1] = 0;
		else if (h[t1 - 1] > h[t2 - 1]) q[t2 - 1] = 0;
		else { q[t1 - 1] = 0; q[t2 - 1] = 0;}
	}
	for (ll nn = 0; nn < n; ++nn) {
		cnt += q[nn];
	}
	cout << cnt << endl;
	return 0;
}