Problem:
Score : 500 points

Problem Statement

For two sequences S and T of length N consisting of 0 and 1, let us define f(S, T) as follows:

Consider repeating the following operation on S so that S will be equal to T. f(S, T) is the minimum possible total cost of those operations.

Change S_i (from 0 to 1 or vice versa). The cost of this operation is D \times C_i, where D is the number of integers j such that S_j \neq T_j (1 \leq j \leq N) just before this change.

There are 2^N \times (2^N - 1) pairs (S, T) of different sequences of length N consisting of 0 and 1. Compute the sum of f(S, T) over all of those pairs, modulo (10^9+7).

Constraints

1 \leq N \leq 2 \times 10^5

1 \leq C_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
C_1 C_2 \cdots C_N

Output

Print the sum of f(S, T), modulo (10^9+7).

Sample Input 1

1
1000000000

Sample Output 1

999999993

There are two pairs (S, T) of different sequences of length 2 consisting of 0 and 1, as follows:

S = (0), T = (1): by changing S_1 to 1, we can have S = T at the cost of 1000000000, so f(S, T) = 1000000000.

S = (1), T = (0): by changing S_1 to 0, we can have S = T at the cost of 1000000000, so f(S, T) = 1000000000.

The sum of these is 2000000000, and we should print it modulo (10^9+7), that is, 999999993.

Sample Input 2

2
5 8

Sample Output 2

124

There are 12 pairs (S, T) of different sequences of length 3 consisting of 0 and 1, which include:

S = (0, 1), T = (1, 0)

In this case, if we first change S_1 to 1 then change S_2 to 0, the total cost is 5 \times 2 + 8 = 18. We cannot have S = T at a smaller cost, so f(S, T) = 18.

Sample Input 3

5
52 67 72 25 79

Sample Output 3

269312

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

program change_a_little_bit
    implicit none
    integer(8),parameter:: md=1000000007_8
    integer(8):: n,i,ans=0
    integer(8),allocatable:: c(:)

    read*,n
    allocate(c(n))
    read*,c(:)
    call merge_sort(c,1_8,n)
    print*, c(:)
    
    do i=1, n
        ans = mod(ans + mod(c(i)*mod(pow_fast(2_8,n-1,md)+mod(pow_fast(2_8,n-2,md)*(n-i),md),md),md),md)
    end do

    print*, mod(ans*pow_fast(2_8,n,md),md)
contains

recursive subroutine merge_sort(ar,fst,lst)
    integer(8):: ar(:), fst, lst, mid, tmp

    if (lst-fst < 2) then
        if(ar(lst) < ar(fst)) then
            tmp = ar(lst)
            ar(lst) = ar(fst)
            ar(fst) = tmp
        end if
        return
    end if

    mid = (fst+lst)/2
    call merge_sort(ar,fst,mid)
    call merge_sort(ar,mid+1,lst)
    call merge(ar,fst,mid,lst)
end subroutine

subroutine merge(ar,fst,mid,lst)
    integer(8):: ar(:), fst, mid, lst
    integer(8),pointer:: tmp(:)
    integer(8):: it,il,ir

    allocate(tmp(lst-fst+1))
    it = 1
    il = fst
    ir = mid+1
    do while(il<=mid .and. ir<=lst)
        if (ar(il) <= ar(ir)) then
            tmp(it) = ar(il)
            il=il+1        
        else
            tmp(it) = ar(ir)
            ir=ir+1
        end if
        it=it+1
    end do

    if (ir > lst) then
        tmp(it:) = ar(il:mid)
    else
        tmp(it:) = ar(ir:lst)
    end if

    ar(fst:lst) = tmp(:)
    deallocate(tmp)
end subroutine

function pow_fast(x,pow,md) result(ret)
    integer(8),value:: x, pow
    integer(8)::md,ret

    ret=1_8
    if (x==1 .or. pow==0) return

    do while(pow > 0_8)
        if (mod(pow,2_8) == 1_8) then
            ret = mod(ret*x,md)
        end if
        x = mod(x*x,md)
        pow = pow/2_8
    end do

end function



end program change_a_little_bit