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 PyPy3 (7.3.0) (judge verdict: Accepted)

ma = lambda :map(int,input().split())
lma = lambda :list(map(int,input().split()))
tma = lambda :tuple(map(int,input().split()))
ni = lambda:int(input())
yn = lambda fl:print("Yes") if fl else print("No")
import collections
import math
import itertools
import heapq as hq
nmax = 2*10**5+1
mod = 10**9+7
fac = [1]*nmax
finv = [1]*nmax
inv = [1]*nmax
def ncr_pre():
    for i in range(2,nmax):
        fac[i] = fac[i-1]*i % mod
        inv[i] = mod - inv[mod%i] * (mod//i) %mod
        finv[i] = finv[i-1] * inv[i] %mod

def ncr(n,r):
    if n<r:
        return 0
    if n<0 or r<0:
        return 0
    return fac[n]* (finv[r] * finv[n-r] %mod) %mod
n = ni()
C = lma();C.sort()
ans = 0
for i in range(1,n+1):
    c = C[i-1]
    ans += c*pow(4,i-1,mod)*pow(2,n-i,mod)*pow(2,n-i,mod)*(n-i+2)
    ans%=mod
print(ans)