Problem:
Score : 400 points

Problem Statement

You are given N items.

The value of the i-th item (1 \leq i \leq N) is v_i.

Your have to select at least A and at most B of these items.

Under this condition, find the maximum possible arithmetic mean of the values of selected items.

Additionally, find the number of ways to select items so that the mean of the values of selected items is maximized.

Constraints

1 \leq N \leq 50

1 \leq A,B \leq N

1 \leq v_i \leq 10^{15}

Each v_i is an integer.

Input

The input is given from Standard Input in the following format:

N A B
v_1
v_2
...
v_N

Output

Print two lines.

The first line should contain the maximum possible arithmetic mean of the values of selected items. The output should be considered correct if the absolute or relative error is at most 10^{-6}.

The second line should contain the number of ways to select items so that the mean of the values of selected items is maximized.

Sample Input 1

5 2 2
1 2 3 4 5

Sample Output 1

4.500000
1

The mean of the values of selected items will be maximized when selecting the fourth and fifth items. Hence, the first line of the output should contain 4.5.

There is no other way to select items so that the mean of the values will be 4.5, and thus the second line of the output should contain 1.

Sample Input 2

4 2 3
10 20 10 10

Sample Output 2

15.000000
3

There can be multiple ways to select items so that the mean of the values will be maximized.

Sample Input 3

5 1 5
1000000000000000 999999999999999 999999999999998 999999999999997 999999999999996

Sample Output 3

1000000000000000.000000
1

Solution in Fortran (gfortran v4.8.4) (judge verdict: Accepted)

program main
  implicit none
  integer(16)::N,A,B
  integer(16),allocatable,dimension(:,:)::C
  integer(16),allocatable,dimension(:)::v
  integer(16)::i
  integer(16),parameter::mo=10**9+7_16
  real(16)::ANS1
  integer(16)::ANS2
  integer(16)::cnt,cnt2
  read*,N,A,B
  allocate(v(N))
  read*,v
  call heapsort(N,v)
  ANS1=0
  do i=N,N-A+1,-1
   ANS1=ANS1+v(i)
  end do
  ANS1=ANS1/real(A)
  print"(f0.20)",ANS1
  call comb_table(N)

  if(v(N)/=v(N-A+1))then
   cnt=count(v(1:n) == v(n-a+1))
   cnt2 = count(v(n-a+1:n) == v(n-a+1))
   ANS2=C(cnt,cnt2)
  else
   cnt=A
   do i=N-A,1,-1
      if(v(i)/=v(N))exit
      cnt=cnt+1
   end do
   ANS2=0
   do i=A,B
      ANS2=ANS2+C(cnt,i)
   end do
  endif
  print"(i0)",ANS2
contains
subroutine heapsort(n,array)
  implicit none
!ここの入力は状況に応じて変更すること
  integer(16),intent(in) :: n
  integer(16),intent(inout) :: array(1:n)
  integer(16)::i,k,j,l
  integer(16):: t

  l=n/2+1
  k=n
  do while(k /= 1)
     if(l > 1)then
        l=l-1
        t=array(L)
     else
        t=array(k)
        array(k)=array(1)
        k=k-1
        if(k == 1) then
           array(1)=t
           exit
        endif
     endif
     i=l
     j=l+l
     do while(j<=k)
        if(j < k)then
           if(array(j) < array(j+1))j=j+1
        endif
        if (t < array(j))then
           array(i)=array(j)
           i=j
           j=j+j
        else
           j=k+1
        endif
     enddo
     array(i)=t
  enddo
  return
end subroutine heapsort
subroutine comb_table(N)
   integer(16)::N
   integer(16)::i,j
   allocate(C(0:N,0:N))
   C=0
   do i=0,N
      do j=0,i
         if(j==0 .or. j==i)then
            C(i,j)=1
         else
            C(i,j)=C(i-1,j-1)+C(i-1,j)
         endif
      end do
   end do
end subroutine
end program main