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>    
//#include <ext/pb_ds/assoc_container.hpp> 
//#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//using namespace __gnu_pbds; 
#define fastio() ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define pb  push_back
#define show(x) cout<<(#x)<<" : "<<x<<endl;
#define ll  long long
#define ld  long double
#define fill(a,val) memset(a,val,sizeof(a))
#define mp  make_pair
#define ff  first
#define ss  second
#define pii pair<ll,ll>
#define sq(x) ((x)*(x))
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(),v.rend()
#define endl "\n"
#define int long long 
#define printclock cerr<<"Time : "<<1000*(ld)clock()/(ld)CLOCKS_PER_SEC<<"ms\n";
const ll MOD     = 1000*1000*1000+7;
const ll INF     = 1ll*1000*1000*1000*1000*1000*1000 + 7;
const ll MOD2    = 998244353;
const ll N       = 1000 * 100 + 10;
const ll N2      = 70;
const ld PI  = 3.141592653589793;
//template<class T> using oset=tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
ll gcd(ll a, ll b){if(!b)return a;return gcd(b, a % b);} 
ll power(ll x,ll y,ll p = LLONG_MAX ){ll res=1;x%=p;while(y>0){if(y&1)res=(res*x)%p;y=y>>1;x=(x*x)%p;}return res;}
ll lcm(ll a , ll b){return (a*b)/gcd(a,b);}
void solve(){
  #ifndef ONLINE_JUDGE    
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
  #endif
}
signed main()
{
    fastio();
    //cout<<fixed<<setprecision(20);
    //CHECK for LONG LONG and LONG DOUBLE
    //*comment for all except cc/cf   
    ll n,m;
    cin>>n>>m; 
    ll h[n];
    for(ll i=0;i<n;i++) 
    cin>>h[i];
    ll a2[100000];
    fill(a2,0);
    for(ll i=0;i<m;i++){
      ll a,b;
      cin>>a>>b; 
      a2[a-1]=max(a2[a],h[b]);
      a2[b-1]=max(a2[b],h[a]);
    }
    ll ans=0;
    for(ll i=0;i<n;i++){
      ans+=(h[i]>a2[i]);
    }
    cout<<ans<<endl; 
    
    return 0;
}