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 <iomanip> //! for setprecision(10)
#include <math.h>
#include <algorithm>
#include <functional> 
#include <string>
#include <vector>

#include <cstring>
#include <set>
#include <map>
#include <queue>
#include <utility>

using namespace std;

typedef long long LL;

#define dump(c) { for (auto it = c.begin(); it != c.end(); ++it) if (it == c.begin()) cout << *it; else cout << ' ' << *it; cout << endl; } 
#define rep(i,n) for (int i = 0; i < (n); ++i)

const int MOD = 1000000007;

int main()
{
	cin.tie(0);
	ios::sync_with_stdio(0);

    int N, M;
    cin >> N >> M;

    vector<int> H(N);
    vector<bool> bad(N);

    rep(i, N) cin >> H[i];
    rep(i, M) {
        int A, B;
        cin >> A >> B;
        --A; --B;
        if (H[A] <= H[B]) bad[A] = true;
        if (H[B] <= H[A]) bad[B] = true;
    }

    int ans = 0;
    rep(i, N) {
        if (!bad[i]) ++ans;
    }

    cout << ans << endl;

    return 0;
}