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 Rust (1.15.1) (judge verdict: Accepted)

use std::ops::Sub;
use std::ops::Div;
use std::ops::Mul;
use std::ops::Rem;
use std::ops::ShrAssign;
use std::ops::BitAnd;

const MOD: i64 = 1000000007;

fn main() {
    let n: i64 = read();
    let mut cn: Vec<i64> = read_vec();
    cn.sort();
    cn.reverse();
    let mut ans = 0;
    for (i, &c) in cn.iter().enumerate() {
        ans += c * (i as i64 + 2) % MOD;
        ans %= MOD;
    }
    ans *= modpow(4, n-1, MOD);
    ans %= MOD;

    println!("{:?}", ans);
}
fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}
fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().split_whitespace()
        .map(|e| e.parse().ok().unwrap()).collect()
}

/// calculate $a^n % m$
/// a > 0, n >= 0, m > 0
pub fn modpow<T>(mut a: T, mut n: T, m: T) -> T
    where T: Sub<Output=T> + Div<Output=T> + Mul<Output=T> + Rem<Output=T> + ShrAssign + PartialOrd + BitAnd<Output=T> + Copy + From<u8>
{
    let zero = T::from(0);
    let one = T::from(1);
    let mut res = one;

    while n > zero {
        if (n & one) == one {
            res = res * a % m;
        }
        a = a * a % m;
        n >>= one;
    }

    res
}