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: Accepted)

#!/usr/bin/env python3
import sys

class Node:
    def __init__(self, h):
        self.high = h
        self.ne_max = 0

    def set_ne(self, h):
        self.ne_max = max(self.ne_max, h)

    def isg(self):
        return 1 * (self.high > self.ne_max)

def solve(N: int, M: int, H: "List[int]", A: "List[int]", B: "List[int]"):
    nodes = [Node(h) for h in H]

    for ai, bi in zip(A,B):
        nodes[ai-1].set_ne(nodes[bi-1].high)
        nodes[bi-1].set_ne(nodes[ai-1].high)

    print(sum([n.isg() for n in nodes]))

    return


# Generated by 1.1.6 https://github.com/kyuridenamida/atcoder-tools  (tips: You use the default template now. You can remove this line by using your custom template)
def main():
    def iterate_tokens():
        for line in sys.stdin:
            for word in line.split():
                yield word
    tokens = iterate_tokens()
    N = int(next(tokens))  # type: int
    M = int(next(tokens))  # type: int
    H = [int(next(tokens)) for _ in range(N)]  # type: "List[int]"
    A = [int()] * (M)  # type: "List[int]"
    B = [int()] * (M)  # type: "List[int]"
    for i in range(M):
        A[i] = int(next(tokens))
        B[i] = int(next(tokens))
    solve(N, M, H, A, B)

if __name__ == '__main__':
    main()