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 <bits/stdc++.h>

#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()

using namespace std;

typedef long long ll;
typedef vector<int> vi;

void solve(){
	
	int n, m; cin >> n >> m;	
	vi H;
	int ans = 0;
	for(int i = 0, tmp; i < n; ++i){
		cin >> tmp;
		H.push_back(tmp);
	}
	vector<vi> adj(n);
	for(int i = 0, tmp1, tmp2; i < m; ++i){
		cin >> tmp1 >> tmp2; tmp1--; tmp2--;
		adj[tmp1].push_back(tmp2);
		adj[tmp2].push_back(tmp1);
	}
	for(int i = 0; i < n; ++i){
		int ok = 1;
		for(int j = 0; j < adj[i].size(); ++j){
			if(H[i] <= H[adj[i][j]]){
				ok = 0;
			}
		}
		ans += ok;
	}
	cout << ans << "\n";	
	
			
}

int main(){
	std::ios::sync_with_stdio(0); std::cin.tie(0);
	
	solve();
		
	return 0;
}