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: Accepted)

#include <cstdio>
#include <vector>

#define rep(i, n) for (int i = 0; i < n; ++i)
#define rep(i, n) for (int i = 0; i < n; ++i)

typedef std::vector<int> v;
typedef std::vector<std::vector<int>> vv;

int main(void)
{
    int n, m;
    scanf("%d %d", &n, &m);

    v hh(n);
    rep(i, n)
    {
        int h;
        scanf("%d", &h);
        hh[i] = h;
    }

    vv ggg(n);

    rep(i, m)
    {
        int a, b;
        scanf("%d %d", &a, &b);
        ggg[a - 1].push_back(b);
        ggg[b - 1].push_back(a);
    }

    v aa(n);
    rep(i, n)
    {
        int p = hh[i];
        for (auto x : ggg[i])
        {
            if (p >= hh[x - 1])
                aa[x - 1] = 1;
        }
    }

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

    printf("%d", ans);
}