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)

//“ In order to be irreplaceable, one must always be different”
#include <bits/stdc++.h>
using namespace std;
# define ll long long
#define endl "\n"
# define pi 3.14159265359
#define mod 1000000007
# define str string
# define vll vector<ll>
# define vstr vector<str>
# define vchar vector<char>
# define pb push_back
# define mll map<ll,ll>
#define mchar map<char, ll>
# define mstr map<str,ll>
# define vp vector< pair<ll,ll> >
#define mp make_pair
#define ff first
#define ss second
#define pll pair<ll,ll>
# define f(x,y) for(int x=0;x<y;x++)
# define r(x,y) for(int x=y;x>=0;x--)
# define loop(x,y,z) for(int x=y;x<z;x++)
ll power(ll a , ll b){ll res=1;for(ll i=0; i<b; ++i){res*=a;}return res;}
ll logx(ll base, ll num){int cnt=0;while(num!=1){num/=base; ++cnt;}return cnt;}
ll gcd(ll a, ll b){
    if(b == 0)
        return a;
    return gcd(b, a % b);
}
ll lcm(ll a, ll b){
    return (a*b)/gcd(a, b);
}
void print(vll a){
    for(int i=0;i<a.size();i++){
        cout<<a[i]<<' ';
    }
    cout<<endl;
}
vll inp(int n){
    vll a;
    ll x;
    f(i,n){
        cin>>x;
        a.pb(x);
    }
    return a;
}

int main(){
    ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
    ll x,y;
    cin>>x>>y;
    ll a[x];
    f(i,x){
        cin>>a[i];
    }
    vll v1;
    v1.resize(x);
    for(ll i=0;i<x;i++){
        v1[i]=0;
    }
    f(i,y){
        ll b,c;
        cin>>b>>c;
        b--;
        c--;
        if(a[b]<a[c]){
            if(v1[b]==0 || v1[b]==1){
                v1[b]=2;
            }
            else{
                if(v1[c]==0){
                v1[c]=1;
                }
            }
        }
        else {
            if(v1[c]==0 || v1[c]==1){
                v1[c]=2;
            }
            else{
                if(v1[b]==0){
                v1[b]=1;
            }
            }
        }
    }
    ll ans=0;
    f(i,v1.size()){
        if(v1[i]==1 || v1[i]==0){
            ans++;
        }
    }
    cout<<ans;
    return 0;
}