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 Python (3.8.2) (judge verdict: Accepted)

import sys

if __name__ == '__main__':
    n, m = list(map(lambda x: int(x), sys.stdin.readline().split(' ')))
    h = list(map(lambda x: int(x), sys.stdin.readline().split(' ')))
    # matrix = {}
    # matrix = [[0 for i in range(n)] for j in range(n)]
    matrix = [[] for j in range(n)]

    for i in range(m):
        a, b = list(map(lambda x: int(x), sys.stdin.readline().split(' ')))
        a, b = a-1, b-1
        # matrix[a] = matrix.get(a, []) + [b]
        # matrix[b] = matrix.get(b, []) + [a]
        # matrix[a][b] = 1
        # matrix[b][a] = 1
        matrix[a].append(b)
        matrix[b].append(a)

    bads = {}
    good = 0
    # 展望台iが良い展望台か
    for i in range(n):
        if bads.get(i, 0) == 1:
            continue
        # for t in matrix.get(i, []):
        for t in matrix[i]:
            if h[i] <= h[t]:
                break
            else:
                # 展望台tはもはや良い展望台でない
                if t > i:
                    bads[t] = 1
        else:
            good += 1
    print(good)

# matrixを疎行列にする
# 判定順をソートする