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: Wrong Answer)

# from collections import defaultdict

n, m = map(int, input().split())
high = [int(i) for i in input().split()]
# a_b = [[int(i) for i in input().split()] for i in range(m)]

high_d = {}
for i in range(n):
    # d = defaultdict(list)
    high_d[i+1] = high[i]
print(high_d)

d = {}
for _ in range(m):
    a, b = map(int, input().split())
    if a not in d:
        d[a] = [b]  # リストにし忘れ
    else:
        d[a].append(b)
    if b not in d:
        d[b] = [a]  # リストにし忘れ
    else:
        d[b].append(a)


print(d)
cnt = 0
for i, li in d.items():
    # [a].append(b)
    is_true = True
    for j in li:
        if high_d[i] > high_d[j]:
            continue
        else:
            is_true = False
    if is_true:
        cnt += 1

cnt += n - len(d)

print(cnt)