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 <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <cmath>
#include <bitset>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <algorithm>
#include <complex>
#include <unordered_map>
#include <unordered_set>
#include <random>
#include <cassert>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
ll par[100002];
ll rk[100002];
ll sz[100002];
void init(ll n){
	for(ll i=0; i<n; i++){
		par[i]=i; sz[i]=1;
	}
}
 
ll find(ll x){
	if(par[x]==x){
		return x;
	}else{
		return par[x]=find(par[x]);
	}
}
 
void unite(ll x, ll y){
	x=find(x);
	y=find(y);
	if(x==y) return;
	if(rk[x]<rk[y]){
		par[x]=y; sz[y]+=sz[x];
	}else{
		par[y]=x; sz[x]+=sz[y];
		if(rk[x]==rk[y]) rk[x]++;
	}
}
 
bool same(ll x, ll y){
	return find(x)==find(y);
}
int main()
{
  ll n, m; cin>>n>>m;
  ll ans[100002];
  ll a[100002], b[100002];
  ll wk;
  for(ll i=0; i<m; i++){
	  cin >> wk;
  }
  for(ll i=0; i<m; i++){
    cin>>a[i]>>b[i];
    a[i]--; b[i]--;
  }
  init(n);
  ll c=0;
  //ll x,y;
  for(ll i=m-1; i>=0; i--){
    ll x=a[i], y=b[i];
    if(same(x, y)){
      continue;
    }
    x=find(x), y=find(y);
    unite(x, y);
  }
  map<ll,ll>mp;
  for(ll i=0; i<n; i++){
    ll x=find(i);
	mp[x]++;
	
  }
  printf("%lld",(ll)mp.size());
	return 0;
}