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: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
#define rep(i,n) for(ll i=0;i<n;++i)
#define ocut cout
#define ouct cout
#define itn int

struct Union{
  vector<ll> par;
  Union(ll a){
  par=vector<ll>(a,-1);
  }
  ll find(ll a){
    if(par[a]<0){
      return a;
    }
    else{
      return par[a]=find(par[a]);
    }
  }
  bool same(ll a,ll b){
    return (find(a)==find(b));
  }
  ll size(ll a){
    return -par[find(a)];
  }
  void unite(ll a,ll b){
    ll c=find(a),d=find(b);
    if(c==d)
      return;
    if(size(c)<size(d)){
      swap(c,d);
    }
    par[c]+=par[d];
    par[d]=c;
  }
};
ll Nagaemon1;
void swap(ll i,ll j){
  Nagaemon1=i;
  i=j;
  j=Nagaemon1;
}
int main(){
  ll n,l,r;
  cin >> n;
  vector<ll> a(n);
  vector<ll> w(1000000001),e(1000000001);
  rep(i,n){
    cin >> a[i];
    l=i-a[i];
    r=i+a[i];
    w[l]++;
    e[r]++;
  }
  ll ans=0;
  rep(i,1000000001){
    ans+=w[i]*e[i];
  }
  cout << ans;
}