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)

#pragma GCC optimize("Ofast")

#include <iostream>
#include <string>
#include <vector>
#include <list>
#include <map>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <numeric>
using namespace std;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int N, M;
    vector<unsigned int> Highs;
    vector<unsigned int> keys;
    vector<unsigned int> values;
    cin >> N >> M;

    Highs.resize(N);
    for (unsigned int i = 0; i < N; i++)
    {
            cin >> Highs[i];
    }

    keys.resize(M);
    values.resize(M);

    for (unsigned int i = 0; i < M; i++)
    {
        cin >> keys[i] >> values[i];
    }

    unsigned int count = 0;
    vector<unsigned int> kuList;
    kuList.reserve(20000);
    for (unsigned int i = 0; i < N; i++)
    {
        // くっついてるリスト作成
        kuList.clear();
        for (unsigned int j = 0; j < M; j++)
        {
            if( keys[j] == i + 1 && find(kuList.begin(),kuList.end(),keys[j]) == kuList.end())
            {
                kuList.push_back(keys[j]);
            }
            else if (values[j] == i + 1 && find(kuList.begin(), kuList.end(), values[j]) == kuList.end())
            {
                kuList.push_back(values[j]);
            }
        }

        bool isis = true;
        for (unsigned int k = 0; k < kuList.size(); k++)
        {
            if (Highs[i] < Highs[kuList[k] - 1])
            {
                isis = false;
                break;
            }
        }
        if (isis)
            count++;
    }

    cout << count / 2 << "\n";

    return 0;
}