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;
typedef long long ll;
ll cnt[100002]={};
int main()
{
    cin.tie(0);
    cin.sync_with_stdio(0);

    ll t=1;
    //cin>>t;
    while(t--)
    {
       ll n,m,c=0;
       cin>>n>>m;
       ll a[n+1];
       for(int i=1;i<=n;i++) cin>>a[i];
       while(m--)
       {
           ll x,y;
           cin>>x>>y;
           if (a[x]>a[y] && cnt[x]!=10) {c++;cnt[x]=10;cnt[y]=max(cnt[y],1ll);}
           else if(a[y]>a[x] && cnt[y]!=10) {c++;cnt[y]=10;cnt[x]=max(cnt[x],1ll);}
           else {
                if(cnt[x]==0) cnt[x]=1;
                if(cnt[y]==0) cnt[y]=1;
                }
       }
       for(int i=1;i<=n;i++)
       {
           if(cnt[i]==0) c++;
       }
       cout<< c;
    }
    return 0;
}