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)

#require 'prime'
#require 'openssl'
MOD=10**9+7;  cnt=0; sum=0; prev=nil; can=true; h=Hash.new(0)
def gs() gets.chomp end
def gi() gets.chomp.to_i end
def gsmi() gets.chomp.split.map(&:to_i) end

def ar2(s1,s2,de=nil) Array.new(s1){Array.new(s2,de)} end
def ar3(s1,s2,s3,de=nil) Array.new(s1){Array.new(s2){Array.new(s3,de)}} end

def desc(ar) ar.sort!{|x,y| y<=>x} end
def min(a,b) a<=b ? a : b end
def max(a,b) a>=b ? a : b end
def sum(ar) ar.inject(:+).to_i end
def ceil(a,b) (a.to_f/b).ceil end

def C(a,b) b==0||a==b ? 1 : (b=a-b if a/2<b;(a-b+1..a).inject(:*)/(1..b).inject(:*)) end
def rui(ar) s=[0]; ar.size.times{|i| s<<s[-1]+ar[i]}; s end
def rev_rui(ar) s=[0]; ar.size.times.reverse_each{|i| s<<s[-1]+ar[i]}; s end

def puts_yesno(b) puts(b ? 'Yes' : 'No') end
def putsend(s) puts s; exit end
def pu(h) begin; require './../nemi_lib/debug.rb'; puts_val(h); rescue LoadError; end; end
########### ( ˘ω˘ )ｽﾔｧ… ###########
n,m=gsmi
hh=gsmi
ab=m.times.map{gsmi}
fl=Array.new(n,1)

ab.each do |a,b|
  if hh[a-1]>hh[b-1]
    fl[b-1]=0
  elsif hh[a-1]<hh[b-1]
    fl[a-1]=0
  else
    fl[a-1]=0
    fl[b-1]=0
  end
end

puts fl.count(1)