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: Runtime Error)

# 各々の要素が何番目に何回数えられるかを考える。
import sys

input = sys.stdin.buffer.readline

n = int(input())
C = list(map(int, input().split()))
C.sort(reverse=True)
mod = 10 ** 9 + 7
N = n + 5
bikkuri = [0] * N
bikkuri[0] = 1
gyaku = [0] * N
for i in range(1, N):
    bikkuri[i] = (i * bikkuri[i - 1]) % mod
gyaku[N - 1] = pow(bikkuri[N - 1], mod - 2, mod)
for i in range(N - 1, 0, -1):
    gyaku[i - 1] = (gyaku[i] * i) % mod


def comb(n, r):
    if r < 0 or n - r < 0:
        return 0
    return (bikkuri[n] * gyaku[r] * gyaku[n - r]) % mod


ans = 0
for i in range(n):
    ans += pow(2, i - 1, mod) * (i + 2) * C[i] * pow(2, n - (i + 1), mod)
    ans %= mod
ans *= pow(2, n, mod)
ans %= mod
print(ans)