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: Accepted)

program main
  implicit none
  integer :: N, M
  integer,allocatable :: H(:), A(:), B(:)
  type neig
     integer, allocatable :: cand(:)
  end type neig
  type(neig), allocatable :: obs(:)
  integer, allocatable :: cnt(:)
  logical, allocatable :: flag(:)
  integer :: i,j,k


  read(*,*) N,M
  allocate(H(N),A(M),B(M))
  allocate(obs(N))
  allocate(cnt(N))
  read(*,*) H(:)
  do i = 1,M
     read(*,*) A(i),B(i)
  end do

  cnt(:) = 0
  do i = 1,M
     cnt(A(i)) = cnt(A(i)) + 1
     cnt(B(i)) = cnt(B(i)) + 1
  end do
  do i = 1,N
     allocate( obs(i)%cand(cnt(i)) )
  end do
  cnt(:) = 0
  do i = 1,M
     cnt(A(i)) = cnt(A(i)) + 1
     obs(A(i))%cand(cnt(A(i))) = B(i)
     cnt(B(i)) = cnt(B(i)) + 1
     obs(B(i))%cand(cnt(B(i))) = A(i)
  end do
  allocate(flag(N)) 
  flag(:) = .true.
  do i = 1,N
     do j = 1,cnt(i)
        if ( H(i) <= H( obs(i)%cand(j) ) ) then
           flag(i) = .false.
           exit
        end if
     end do
  end do
  write(*,'(i0)') count(flag)
  stop
end program main