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;
int main(){
  int n,m; cin>>n>>m;
  vector<vector<int>> road(n,vector<int>(n,0));
  vector<int> h(n);
  vector<bool> flag(n,true);
  for(int i=0;i<n;i++) cin>>h[i];
  int tmp_a,tmp_b;
  for(int i=0;i<m;i++){
    cin>>tmp_a>>tmp_b;
    /*
    road[tmp_a-1][tmp_b-1]=1;
    road[tmp_b-1][tmp_a-1]=1;
    */
    if(h[tmp_a-1]>h[tmp_b-1]) flag[tmp_b-1]=false;
    if(h[tmp_a-1]<h[tmp_b-1]) flag[tmp_a-1]=false;
    if(h[tmp_a-1]==h[tmp_b-1]) {flag[tmp_b-1]=false;flag[tmp_a-1]=false;}
  }
  int ans=0;
  //bool flag=true;
  for(int i=0;i<n;i++){
    if(flag[i]){ans++;}
  }
  cout<<ans<<endl;
}