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 ll long long
#define pii pair<int, int>
#define pll pair<long long, long long>
#define pci pair<char, int>
#define pic pair<int, char>
#define plvl pair<long long, vector<long long> >
#define pcvc pair<char, vector<char> >


using namespace std;

const int mod = 1e9 + 7;

ll gcd(ll a, ll b) {
	if(b == 0) {
		return a;
	}
	return gcd(b, a % b);
}

template <typename T>
void show1(vector<T> dp) {
	for(auto d : dp) {
		cout << d << " ";
	}
	cout << endl;
}

template <typename T>
void show2(vector<vector<T> > dp) {
	for(auto d : dp) {
		for(auto dd : d) {
			cout << dd << " ";
		}
		cout << endl;
	}
}

void executeCode() {
	int n, m;
	cin >> n >> m;
	unordered_map<int, vector<int> > mp;
	vector<int> h(n);
	for(int i = 0 ; i < n ; i++) {
		cin >> h[i];
	}
	for(int i = 0 ; i < m ; i++) {
		int a, b;
		cin >> a >> b;
		mp[a].push_back(b);
		mp[b].push_back(a);
	}
	int cnt = 0;
	for(auto m : mp) {
		int mh = -1;
		// cout << m.first << " -- ";
		for(auto mm : m.second) {
			mh = max(mh, h[mm - 1]);
			// cout << mm << ", " << mh << "     ";
		}
		// cout << endl;
		if(mp[m.first].size() == 0 || h[m.first - 1] > mh) {
			cnt++;
		}
	}
	cout << cnt + n - mp.size() << endl;
}

int main() {
	ios::sync_with_stdio(false);
    cin.tie(nullptr);
	int tt = 1;
	// cin >> tt;
	while(tt--) {
		executeCode();	
	}
	return 0;
}