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)

n,m=map(int,input().split())
a=input().split()
l=[]
count=0
for i in a:
  l.append(int(i))
nt=[]
for i in range(1,n+1):
  nt.append(i)
t=[]
ans=[]
f=[]
for j in range(m):
  inp=input().split()
  if (inp[0],inp[1]) not in t:
    t.append((inp[0],inp[1]))
    b=l[int(inp[0])-1]
    c=l[int(inp[1])-1]
    if b>c:
      if inp[1] not in f:
        f.append(inp[1])
      if inp[0] not in f:
        ans.append(inp[0])
      if inp[1] in ans:
        ans.remove(inp[1])
    elif b<c:
      if inp[0] not in f:
        f.append(inp[0])
      if inp[1] not in f:
        ans.append(inp[1])
      if inp[0] in ans:
        ans.remove(inp[0])
    else:
      if inp[0] not in ans and inp[0] not in f:
        ans.append(inp[0])
      if inp[1] not in ans and inp[1] not in f:
        ans.append(inp[1])
  else:
    count+=1
  if int(inp[0]) in nt:
    nt.remove(int(inp[0]))
  if int(inp[1]) in nt:
    nt.remove(int(inp[1]))
ans=set(ans)
print(len(ans)-count+len(nt))