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: Accepted)

#include <iostream>
#include <string>
#include <algorithm>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <iomanip>
#include <bitset>

#define rep(i,n) for (int i = 0; i < (n); i++)
#define FOR(i,a,b) for (int i=(a); i < (b); i++)
#define INF 100000000000
#define MOD 100000007 //10^9+7
using namespace std;
using ll = long long;
using P = pair<int, int>;

char s[50][50];
int h, w;

int main()
{
    int n, m;
    cin >> n >> m;
    vector<int> h(n);
    rep(i,n) cin >> h[i];

    vector<vector<int>> arr(n);
    int a, b;
    rep(i,m) {
        cin >> a >> b;
        a--; b--;
        arr[a].push_back(b);
        arr[b].push_back(a);
    }

    int ans = 0;
    for (int i = 0; i < n; i++) {
        int flag = 1;
        if (arr[i].size() == 0 ) {
            ans++;
            continue;
        }
        for (int j = 0; j < arr[i].size(); j++) {
            if (h[i] <= h[arr[i][j]]) {
                flag = 0;
            }
        }
        if (flag)
            ans++; 
    }
    cout << ans << endl;
         
    return 0;
}