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)

/**-------------------------
|--------From A.R.A--------|
-------------------------**/
#include <bits/stdc++.h>

#define ll long long int
#define ull unsigned long long int
#define FOR(i,n) for(ll i=0;i<n;i++)
#define input (a,i,n)for(ll i=0;i<n;i++){      cin>>a[i];}
#define output(a,i,n)for(ll i=0;i<n;i++){cout<<a[i]<<" ";}
#define pb push_back
#define sz(c) (c).size()
#define vi vector <int>
#define vs vector <string>
#define vc vector <char>
#define vl vector <ll>
#define sall(c) sort((c).begin(),(c).end())
#define rsall(c) sort((c).rbegin(),(c).rend())
#define max3(a,b,c) max(max(a,b),c)
#define min3(a,b,c) min(min(a,b),c)

int64_t TN=1;

using namespace std;
void  solution(){
    int n,m,q=0,x,y,l;
    cin>>n>>m;
    int a[n];
    set<int> s,k;
    for(int i=1;i<=n;i++){cin>>a[i];}
    for(int i=1;i<=m;i++){
        cin>>x>>y;
        s.insert(x);s.insert(y);
        if(x!=y){
            l=max(a[x],a[y]);
            k.insert(l);
        }
    }
    cout<<k.size()+n-s.size();

}
int main()
{
    //cin>>TN;
    while(TN--){
        solution();
    }
    return 0;
}