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>
using ll=long long;
using namespace std;
ll INF=1000000000000000001;
ll mod=1000000007;
 
ll i,j,k;
 
int main(){
    ll n,m;cin>>n>>m;
    vector<ll> h(n),a(m),b(m);
    vector<ll> ans(n,-1);
    vector<vector<ll>> G(n);
 
    for(i=0;i<n;i++){cin>>h[i];}
 
    for(i=0;i<m;i++){
        cin>>a[i]>>b[i];
        a[i]--;b[i]--;
 
        G[a[i]].push_back(b[i]);
        G[b[i]].push_back(a[i]);
    }//a,bが展望台を結ぶ
 
    
 
    ll answer=0;
    for(i=0;i<n;i++){
        if(ans[i]!=-1) continue;//すでにvisit
 
        ll max_h=h[i],max_h2=0;
        answer++;
        ans[i]=1;//iをvisit
 
        queue<ll> que;
        que.push(i);
        while(que.size()!=0){
            ll now=que.front();
            que.pop();
 
            for(auto next:G[now]){
                if(ans[next]!=-1) continue;//訪れてしまっている
                
                que.push(next);
                ans[next]=1;
 
                if(max_h!=h[next]){max_h=max(max_h,h[next]);}
                else{max_h2=max(max_h2,h[next]);}
            }
        }
        if(max_h==max_h2 && max_h!=0){answer--;}
    }
 
    
 
    cout<<answer+1<<endl;
    return 0;
}