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<bits/stdc++.h>
#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<cmath>
#include<map>
#include<set>
using namespace std;
typedef long long ll;

#define INF 1000000000000
#define MOD 10000007
//#define int long long
#define pb push_back

void YesNo(bool b) { b ? cout << "Yes" << endl : cout << "No" << endl; return; }
void YESNO(bool b) { b ? cout << "YES" << endl : cout << "NO" << endl; return; }
void yesno(bool b) { b ? cout << "yes" << endl : cout << "no" << endl; return; }

int main() {
	int n, m;
	cin >> n >> m;
	bool yama[100000];
	int h[100000];
	int a, b;
	int dddd;
	int ans = 0;
	vector<int>bbbb[100000];
	for (int i = 0; i < n; i++)
	{
		yama[i] = 0;
		cin >> h[i];
	}
	for (int i = 0; i < m; i++)
	{
		cin >> a >> b;
		bbbb[a - 1].push_back(b);
	}
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < bbbb[i].size(); j++)
		{
			if (h[i]>h[bbbb[i][j]])
			{
				yama[bbbb[i][j]] = false;
			}
		}
	}
	for (int i = 0; i < n; i++)
	{
		if (yama[i])
		{
			ans++;
		}
	}
	cout << ans << endl;
}