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>
#include<string.h>
#include<vector>
#include<set>
#include<map>
#define endl  "\n"
#define ll long long
#define int long long
#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);cin.exceptions(cin.failbit);cout.tie(NULL);
using namespace std;
bool isPrime(int n)
{
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
    if (n % 2 == 0 || n % 3 == 0)
        return false;
    for (int i = 5; i * i <= n; i += 6)
    {
        if ((n % i == 0) || (n % (i + 2) == 0))
            return false;
    }
    return true;
}
vector<int> raj;
void dfs(int vis[],vector<int> v[],int curr,int a[],int n)
{
	vis[curr]=1;
	int i=curr;
	//cout << curr << " ";
	raj.push_back(curr);
	
	for(int j=0;j<v[i].size();j++)
	{
		int x=v[i][j];
		if(vis[x]==0)
		{
			
			dfs(vis,v,x,a,n);
		}
	}
}
int32_t main()
{
	int n,m;
	cin >> n >> m;
	int a[n+1];
	for(int i=1;i<=n;i++)
	{
		cin >> a[i];
	}	
	vector<int> v[n+1];
	for(int i=0;i<m;i++)
	{
		int x,y;
		cin >> x >> y;
		v[x].push_back(y);
		v[y].push_back(x);
	}
	int cnt=0;
	for(int i=1;i<=n;i++)
	{
		int temp=0;
		for(int j=0;j<v[i].size();j++)
		{
			int x=v[i][j];
			temp=max(temp,a[x]);
			
		}
		if(a[i]>temp)
		cnt++;
	}	
	cout << cnt;
}