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 Ruby (2.7.1) (judge verdict: Accepted)

N, M = gets.split.map(&:to_i)
h = []
h = gets.split.map(&:to_i)
a = []
b = []
"想定解"
# ma = []
# ma = ma.fill(0,0..N-1)
# M.times do |i|
#   a[i], b[i] = gets.split.map(&:to_i)
#   ma[a[i]-1] = [ h[b[i]-1], ma[a[i]-1] ].max
#   ma[b[i]-1] = [ h[a[i]-1], ma[b[i]-1] ].max
# end
# ans = 0
# N.times do |i|
#   if ma[i] < h[i]
#     ans += 1
#   end
# end
# puts ans

"隣接リストを用いる"
Adj = Struct.new(:id, :l)
adj = []
N.times do |i|
  adj[i] = Adj.new(i, [])
end
  
M.times do |i|
  a[i], b[i] = gets.split.map(&:to_i)
  adj[a[i]-1].l.push(b[i]-1)
  adj[b[i]-1].l.push(a[i]-1)
end

# puts "#{adj}"

ans = 0
adj.each do |n|
  if n.l.empty?
    ans += 1
  else
    x = 0
    x = n.l.map {|v| v = h[v]}.max
    # puts "n #{n.id}"
    # puts "x #{x}"
    if x < h[n.id]
      ans += 1
    end
  end
end
puts "#{ans}"