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)

import numpy

inp=input().split(" ")
n=int(inp[0])
m=int(inp[1])
inh=input().split(" ")
for i in range(len(inh)):
  inh[i]=int(inh[i])
pairs=numpy.zeros(2*m,dtype=">i4").reshape(-1,2)
for i in range(m):
  a=input().split(" ")
  pairs[i][0]=int(a[0])
  pairs[i][1]=int(a[1])
lists=[[] for i in range(n+1)]
for i in range(len(pairs)):
  lists[pairs[i][0]-1].append(pairs[i][1]-1)
  lists[pairs[i][1]-1].append(pairs[i][0]-1)
for i in range(len(lists)):
  lists[i]=list(set(lists[i]))
count=0
for i in range(len(lists)):
  if len(lists[i])==0:
    count+=1
    print(i)
    continue
  for j in range(len(lists[i])):
    if inh[i] <= inh[lists[i][j]]:
      break
    if j == len(lists[i]):
      count+=1
      print(i)
print(count)