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: Wrong Answer)

#include<bits/stdc++.h>

using namespace std;

#define ld long double
#define ll long long int
#define endl '\n'
#define check_time() cerr<<"Time : "<<1000*(ld)clock()/(ld)CLOCKS_PER_SEC<<"ms\n";
#define fastio() ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define MOD 1e9+7
#define INF 0x3f3f3f3f
#define maxarr(A,n) *max_element(A,A+n) 
#define minarr(A,n) *min_element(A,A+n);

int main()
{
	fastio();
	int N,M,a,b,i;
	ll ans=0;
	cin>>N>>M;
	int H[N+1];
	int Noroads[N+1]={};

	set<int> s;

	for(i=0;i<N;i++)
		cin>>H[i+1];
	for(i=0;i<M;i++)
	{
		cin>>a>>b;
		Noroads[a]++;
		Noroads[b]++;
		if(H[a]>H[b])
		{
			s.insert(a);
			auto pos= s.find(b);
			if(pos!=s.end())
				s.erase(b);
		}
		else if(H[b]>H[a])
		{
			s.insert(b);
			auto pos= s.find(a);
			if(pos!=s.end())
				s.erase(a);
		}
	}

	for(i=1;i<N+1;i++)
	{
		if(Noroads[i]==0)
			ans++;
	}
	ans+=s.size();
	cout<<ans<<endl;
	return 0;   
}