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: Time Limit Exceeded)

N, M = map(int, input().split())  #数字
H = list(map(int, input().split()))#リスト
A = [list(map(int, input().split())) for i in range(M)]  #数字、複数行複数列

good = []
bad = []
for i in range(M):
    left = H[A[i][0] - 1]
    right = H[A[i][1] - 1]
    if left < right:
        good.append(A[i][1])
        bad.append(A[i][0])
    if left > right:
        good.append(A[i][0])
        bad.append(A[i][1])
    if left == right:
        bad.append(A[i][0])
        bad.append(A[i][1])

B = list(i for i in set(good))
C = list(i for i in set(bad))

#print(B)
#print(C)

D = list(i for i in B if i not in C)
E = D + C
F = list(i for i in range(1,N+1) if i not in E)

#print(D)
#print(F)
print(len(D) + len(F))