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>
#define ll long long
#define ld long double
#define mp make_pair
#define pb push_back
#define vll vector<ll>
#define endl "\n"
#define pll pair<ll,ll>
#define all(x) (x).begin() , (x).end()
#define f first
#define s second
#define pr(x) cout<<x<<endl;
#define pr2(x,y) cout<<x<<" "<<y<<endl;
#define pr3(x,y,z) cout<<x<<" "<<y<<endl;
#define prv(v) for(auto x:v) cout<<x<<" ";
#define ffs fflush(stdout);
#define int ll
using namespace std;
 
const ll N =(2e5+5);
const ll MOD = 1e9+7;
const ll INF = LLONG_MAX;
const ll LOG = 29;
#define PI 3.141592653589793238 


long long binpow(long long a, long long b) {
    a%=MOD;    
    long long res = 1;
    while (b > 0) {
        if (b & 1)
            res = (res * a)%MOD;
        a = (a * a)%MOD;
 
        
        b >>= 1;
    }
    res%=MOD;
    return res;
}
void solve(){
    ll n,m;
    cin>>n>>m;
    ll h[n+1];
    for(int i =1;i<=n;i++){
        cin>>h[i];
    }
    vll adj[n+1];
    for(int i=1;i<=m;i++){
        ll u,v;
        cin>>u>>v;
        adj[u].pb(v);
        adj[v].pb(u);
    }
    ll ans = 0;
    for(int i = 1;i<=n;i++){
        bool poss=true;
        for(auto u:adj[i]){
            poss &= (h[i] > h[u]);
        }
        if(poss)ans++;
    }
    cout<<ans<<endl;
}  

 
 
signed main(){
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
     
    ll t=1;
    while(t--){
        solve();
    }    
}