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 PLL pair<ll,ll>
#define F first
#define S second
#define MP make_pair
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
#define N 100001

int main()
{
    fastio;
   
    ll r,i,n;
    cin >> n>>r;
    ll h[n+1];
    for(i=1;i<=n;i++)
    cin >> h[i];
    ll ng[n+1]={0};
    
    while(r--)
    {
        ll u,v;
        cin >> u >> v;
        
        if(h[u]>h[v])
        ng[v]=1;
        
        if(h[v] > h[u])
        ng[u]=1;
        
        if(h[v]==h[u])
        {
            ng[u]=1;
            ng[v]=1;
        }
    }
    ll count=0;
    for(i=1;i<=n;i++)
    {
        if(!ng[i])
        count++;
    }
    
    cout << count <<'\n';
}