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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

class Main {
	Scanner sc;
	int N;
	int[] C;
	static final int MOD = 1000_000_007;
	
	private void calc() {
		sc = new Scanner(System.in);
		N = sc.nextInt();
		C = new int[N];
		for (int i = 0; i < N; i++) C[i] = sc.nextInt();
		Arrays.sort(C);
		long p = 1;
		for (int i = 0; i < N-2; i++) p = p * 2 % MOD;
		long pp = p * 2 % MOD * 2 % MOD;
		
		long a = p*2%MOD;
		long sum = 0;
		long d = (N==1)?0:p;
		for (int i = 0; i < N; i++) {
			sum = (a*C[N-1-i]%MOD + sum)%MOD;
			a = (a + d)%MOD;
		}
		sum = sum * pp % MOD;
		
		System.out.println(sum);
	}
	
	public static void main(String[] args) {
		new Main().calc();
	}
}