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 fio ios_base::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL);
#define Mod 1000000007
#define ll long long
#define dd double
#define pb push_back
#define ff first
#define ss second
#define Mp make_pair
#define INF 999999999999999999
#define NN (ll)(1e6+5)
int mod(string s,int a) 
{ 
    int ans=0; 
    for (int i=0;i<s.length();i++) 
         ans=(ans*10+(int)s[i]-'0')%a; 
    return ans; 
}
ll min(ll x,ll y){if(x<y) return x;return y;}
ll max(ll x,ll y){if(x>y) return x;return y;}
ll power(ll x,unsigned ll y) 
{ 
    if (y==0) 
        return 1; 
    else if (y%2==0) 
        return power(x,y/2)*power(x,y/2); 
    else
        return x*power(x,y/2)*power(x,y/2); 
}
ll hcf(ll a,ll b) 
{ 
    if(b==0) 
        return a; 
    return hcf(b,a%b);  
}
int main()
{
   fio;
   /*#ifndef ONLINE_JUDGE 
    	freopen("input.txt", "r", stdin); 
    	freopen("output.txt", "w", stdout); 
    	freopen("error.txt", "w", stderr); 
   #endif*/
   ll TT=1;
   //cin>>TT;
   while(TT--)
   {    
       ll n,m,i,j,x,y,ans=0;
       cin>>n>>m;
       vector <ll> h(n+1);
       for(i=1;i<=n;i++)
        cin>>h[i];
       vector <ll> v[n+1];
       for(i=0;i<m;i++)
       {
           cin>>x>>y;
           v[x].pb(y);
           v[y].pb(x);
       } 
       for(i=1;i<=n;i++)
       {
           ll c=0;
           for(auto u:v[i])
            if(h[i]<=h[u])
             c++;
           if(!c)
           ans++;  
       }
       cout<<ans<<"\n";
   }
}