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>

using namespace std;

#define INF 1000000007
#define LINF (1LL << 62)
#define PI 3.14159265358979

typedef long long i64;
typedef pair<i64,i64> P;

inline i64 mod(i64 a, i64 m) { return (a % m + m) % m; }

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }



int n, m, a[101010];
vector<int> v[101010];
void solve(){
	cin >> n >> m;
	for(int i = 0; i < n; i++){
		cin >> a[i];
	}
	for(int i = 0; i < m; i++){
		int x,y;
		cin >> x >> y;
		x--; y--;
		v[x].push_back(y);
		v[y].push_back(x);
	}
	
	int cnt = 0;
	for(int i = 0; i < n; i++){
		bool ok = 1;
		for(auto j : v[i]){
			if(a[i] > a[j]) continue;
			ok = 0;
		}
		if(ok) cnt++;
	}
	cout << cnt << endl;
}

int main(){
	std::cin.tie(0);
	std::ios::sync_with_stdio(false);
	int t = 1;
	//cin >> t;
	while(t--){
		solve();
	}
	
	return 0;
}