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 Python (3.4.3) (judge verdict: Accepted)

def examA():
    N = I()
    X = [LI()for _ in range(N)]
    l = 0; r = inf
    while(r-l>_ep):
        now = (l+r)/2
        x_l = -inf; x_r = inf
        y_l = -inf; y_r = inf
        for x,y,c in X:
            move = now / c
            x_l = max(x_l,x-move)
            x_r = min(x_r,x+move)
            y_l = max(y_l,y-move)
            y_r = min(y_r,y+move)
        if x_l<=x_r-_ep and y_l<=y_r-_ep:
            r = now
        else:
            l = now
    ans = r
    print(ans)
    return

def examB():
    N = I()
    C = LI()
    C.sort()
    base = pow(2,2*N-1,mod)
    inv = pow(2,mod-2,mod)
    ans = 0
    for i in range(N):
        cur = C[i]*base*(1+(N-1-i)*inv)
        ans += cur
        ans %= mod
    print(ans)
    return

def examC():
    ans = 0
    print(ans)
    return


import sys,bisect,itertools,heapq,math,random
from copy import deepcopy
from heapq import heappop,heappush,heapify
from collections import Counter,defaultdict,deque
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
def I(): return int(readline())
def LI(): return list(map(int,readline().split()))
def LSI(): return list(map(str,sys.stdin.readline().split()))
def LS(): return sys.stdin.readline().split()
def SI(): return sys.stdin.readline().strip()
global mod,mod2,inf,alphabet,_ep
mod = 10**9 + 7
mod2 = 998244353
inf = 10**18
_ep = 10**(-9)
alphabet = [chr(ord('a') + i) for i in range(26)]

sys.setrecursionlimit(10**7)

if __name__ == '__main__':
    examB()

"""
142
12 9 1445 0 1
asd dfg hj o o
aidn
"""