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 C# (Mono 4.6.2.0) (judge verdict: Time Limit Exceeded)

using System;

namespace ABC150_E___Change_a_Little_Bit
{
    class Program
    {
        static void Main(string[] args)
        {
            var N = int.Parse(Console.ReadLine());
            var C = Array.ConvertAll(Console.ReadLine().Split(' '), int.Parse);
            Array.Sort(C);
            long MOD = 1000000007;
            Combination.combInit(1000000, MOD);

            var result = 0L;
            for (int i = 0; i < N; i++)
            {
                for (int k = 0; k <= N - 1 - i; k++)
                {
                    result += C[i] * power(2, i) * Combination.combination(N - i - 1, k) * (k + 1);
                    result %= MOD;
                }
            }
            result *= power(2, N);
            result %= MOD;
            Console.WriteLine(result);
        }
        static long power(long a, long b)
        {
            if (b == 0) return 1;
            var result = a;
            for (int i = 1; i < b; i++)
            {
                result *= a;
            }
            return result;
        }
    }

    static class Combination
    {
        static long combMax;
        static long combMOD;
        static long[] factorial;
        static long[] inverse;
        static long[] factInv;

        public static void combInit(long _max, long _mod)
        {
            combMax = _max;
            combMOD = _mod;
            factorial = new long[combMax];
            inverse = new long[combMax];
            factInv = new long[combMax];

            factorial[0] = factorial[1] = 1;
            factInv[0] = factInv[1] = 1;
            inverse[1] = 1;

            for (int i = 2; i < combMax; i++)
            {
                factorial[i] = factorial[i - 1] * i % combMOD;
                inverse[i] = combMOD - inverse[combMOD % i] * (combMOD / i) % combMOD;
                factInv[i] = factInv[i - 1] * inverse[i] % combMOD;
            }
        }

        public static long combination(int n, int k)
        {
            if (n < k) return 0;
            if (n < 0 || k < 0) return 0;
            return factorial[n] * (factInv[k] * factInv[n - k] % combMOD) % combMOD;
        }
    }
}