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 PyPy3 (7.3.0) (judge verdict: Wrong Answer)

import sys
from math import sqrt
from collections import Counter, defaultdict, deque

input = sys.stdin.readline
sys.setrecursionlimit(10 ** 6)


def I():
    return int(input())


def MI():
    return map(int, input().split())


def LI():
    return list(MI())


def LIN(n: int):
    return [I() for _ in range(n)]


inf = float("inf")
mod = 10 ** 9 + 7


def main():
    n, m = MI()
    h = LI()

    dict = defaultdict(list)
    for _ in range(m):
        a, b = MI()
        dict[a].append(b)
        dict[b].append(a)

    towers = list(range(1, n + 1))
    for k, v in dict.items():
        if v == []:
            towers.remove(k)
        else:
            for t in v:
                if h[t - 1] >= h[k - 1]:
                    towers.remove(k)
                    break

    print(n - len(towers))


if __name__ == "__main__":
    main()