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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <math.h>
#include <list>
#include <string>
#include <map>
#include <vector>
#include <stack>
#include <queue>
using namespace std;
#define REPD(i, a, b) for (int i = (a); i < (b); ++i)
#define REP(i, n) REPD(i, 0, n)
typedef long long ll;
#define fin(ans) cout << (ans) << '\n'
/*

*/
int main()
{
    ll N,M;
    cin >> N >> M;
    ll H[100100];
    ll cnt = 0;
    //int AB[10010][10010] = {0};
    //bool AB[100][100] = {false};
    map<ll,int> matrix;
    int flags[100100];
    ll temp[2];
    //1~Nまでに入れた
    for(int i=1; i < N+1;i++){
        cin >> H[i];
    }
    REP(i,M){
        cin >> temp[0] >> temp[1];
        matrix[temp[0] * 100000 + temp[1]] = 1;
    }
    ll first,second;
    REPD(i,1,N+1){
        flags[i] = 1;
    }
    for (const auto &[key, value] : matrix)
    {
        first = key%100000;
        second = key/100000;
        //cout << first << " " << second << endl;
        //if(first > second){break;}
        if(H[first] < H[second]){
            //cout << "!" << endl;
            flags[first] = 0;
        }else if(H[second] < H[first]){
            flags[second] = 0;
        }else{
            flags[first] = flags[second] = 0;
        }
        //std::cout << key << " => " << value << "\n";
    }
    REPD(i,1,N+1){
        //cout << flags[i] << endl;
        if(flags[i]) cnt++;
    }
    fin(cnt);
    return 0;
}