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: Runtime Error)

N, M = map(int, input().split())
H_list = list(map(int, input().split()))
A_list = []
B_list = []
for i in range(M):
  A, B = map(int, input().split())
  A_list.append(A)
  B_list.append(B)

count = 0
NG_list = []
for i in range(M):
  NG_list.append(0)
  
while(A_list != NG_list):
  H_max = max(H_list)
  #print(H_list)
  #print("max",H_max)
  max_index = H_list.index(H_max) + 1
  #print("before")
  #print(A_list)
  #print(B_list)
  for i in range(M):
    if((A_list[i] == max_index)or(B_list[i] == max_index)):
      count = count + 1
      B_list[i] = 0
      A_list[i] = 0
  H_list[max_index-1] = 0
  #print("after")
  #print(A_list)
  #print(B_list)
  #print("counter",counter)

print(counter)