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 ll long long
#define io ios_base::sync_with_stdio(false); cin.tie(NULL);
#define pb push_back
#define endl '\n'
#define mod 998244353
#define inf -1000000000000000000
int n,k;
// vector<int>a;
const int sz = 100005;
vector<int>graph[sz];
vector<int>h, p;
int ans = 0;

/*void dfs(int root, vector<bool>&vis) {
	vis[root] = true;
	bool flag = true;
	for(int c : graph[root]) {
		if(!vis[c]) {
			p[c] = root;
			if(h[c] >= h[root]) {
				flag = false;
			}
			dfs(c, vis);
		}
	}
	if(p[root] != -1) {
		if(flag && h[root] > h[p[root]]) ans++;
	}
	else {
		if(flag) ans++;
	}
}*/
int main() {
	io
	cin>>n>>k;
	h.resize(n+1, -1);
	p.resize(n+1, -1);
	for(int i = 1; i<=n; i++) {
		cin>>h[i];
	}
	for(int i = 0; i<k; i++) {
		int a, b;
		cin>>a>>b;
		graph[a].pb(b);
		graph[b].pb(a);
	}
	/*vector<bool>vis(n+1, false);
	for(int i = 1; i<=n; i++) {
		if(!vis[i]) {
			dfs(i, vis);
		}
	}*/
	for(int i = 1; i<=n; i++) {
		if(graph[i].size() == 0) ans++;
		else {
			bool flag = true;
			for(int j = 0; j<graph[i].size(); j++) {
				if(h[i] <= h[graph[i][j]]) flag = false;
			}
			if(flag) ans++;
		}
	}
	cout<<ans;
}