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<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<string>
#include<queue>
#include<set>
#include<map>
using namespace std;
typedef long long int ll;
typedef vector<pair<ll,ll> > Q;
typedef pair<int,int> P;
#define rep(i,j,n) for(int i=(int)(j);i<(int)(n);i++)
const ll inf=1e9+7;
ll gcd(ll a, ll b) { return (b ? gcd(b, a % b) : a); }
ll lcm(ll a, ll b) { return a / (gcd(a, b)) * b; }

int main(){
  int n,m;
  cin>>n>>m;
  vector<int> h(100000);
  rep(i,0,n) cin>>h[i];
  Q ab(100000);
  rep(i,0,m) cin>>ab[i].first>>ab[i].second;

  vector<bool> v(100000);
  rep(i,0,n) v[i]=true;

  rep(i,0,m){
    int a=ab[i].first,b=ab[i].second;
    if(h[a-1]<h[b-1]) v[a-1]=false;
    else if(h[a-1]>h[b-1]) v[b-1]=false;
    else{
      v[a-1]=false,v[b-1]=false;
    }
  }
  int cnt=0;
  rep(i,0,n){
    if(v[i]) cnt++;
  }

  cout<<cnt<<endl;
}