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 Fortran (GNU Fortran 9.2.1) (judge verdict: Time Limit Exceeded)

program peaks

 implicit none
 integer::n,m,i,j,answer,hikaku
 integer,allocatable::h(:),a(:),b(:),ans(:)

 read*,n,m
 allocate(h(0:n),a(m),b(m),ans(n))
  
 h(0)=-1
 read*,h(1:n)
 !print*,h
 do i=1,m
  read*,a(i),b(i)
  !print*,a(i),b(i)
 end do

 ans=0
 answer=0
  
  do i=1,n
   do j=1,m 
   if(i==a(j)) hikaku=b(j)
   if(i==b(j)) hikaku=a(j)
   if(i/=a(j) .and. i/=b(j)) hikaku=0
    if(h(i)<=h(hikaku)) then
    ans(i)=1
    exit
    end if  
   end do
   if(ans(i)==0) answer=answer+1
  end do

  print*,answer

end program peaks