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 ld long double
#define pb push_back
 
#define mo 1000000007
#define inf 1e18
#define rep(i, s, n) for (ll i = s; i < n; i = i + 1)
#define rrep(i,s,n) for(ll i=s;i>=n;i--)
#define adiii \                               
  ios_base::sync_with_stdio(false); \                                      
  cin.tie(NULL);  \
  cout.tie(NULL);									
 
ll mod(ll n) { return (n % (ll)mo + (ll)mo)%(ll)mo;}
ll gcd(ll a,ll b) {if (b == 0) return a;return gcd(b, a % b);}

int main(){
	adiii
	ll n,m;cin>>n>>m;
	vector <ll> h(n+1);
	vector <ll> adj[n+1];
	vector <bool> mark(n+1,false);
	vector <bool> check(n+1,false);
	ll ans = 0;
	rep(i,1,n+1) cin>>h[i];
	ll u,v;
	rep(i,0,m){
		cin>>u>>v;

		if(h[u]>h[v]) adj[u].pb(v);
		else check[u] = true;

		if(h[v]>h[u]) adj[v].pb(u);
		else check[v] = true;

		mark[u] = true;
		mark[v] = true;

	}
	rep(i,1,n+1){
		//cout<<adj[i].size()<<" "<<check[i]<<" "<<mark[i]<<endl;
		
		if(adj[i].size()!=0 && check[i] == false) ans++;
		if(mark[i]==false) ans ++;
		
	}

	cout<<ans<<endl;
}