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>
#define pb push_back
#define rep(i,n) for(int i = 0;i < (n); ++i)
#define all(v) v.begin(),v.end()
#define sort_1(v) sort(v.begin(),v.end())
#define sort_2(v) sort(v.begin(),v.end(),greater<ll>())
typedef long long ll;
typedef long double la;
using namespace std;
 
int main(){
    int n,m;
    cin >> n >> m;
    map<int, int> mp;
    vector<ll> h(n+1);
    for(int i=1;i<n+1;i++) cin >> h[i];
    vector<int> a(n),b(n);
    rep(i,m){
        cin >> a[i] >> b[i];
    }
    rep(i,m){
        if(h[a[i]] > h[b[i]]){
            mp[b[i]] = 1;
        }
        if(h[a[i]] < h[b[i]]){
            mp[a[i]] = 1;
        }
        if(h[a[i]] == h[b[i]]){
            if(mp[a[i]] == 0){
                mp[a[i]] = 1;
            }
            if(mp[b[i]] == 0){
                mp[b[i]] = 1;
            }
        }
    }
    cout << n - mp.size() << endl;
    return 0;
}