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: Accepted)

using System;
using System.Collections.Generic;
using static Assistant.Input;
using System.Linq;
using Assistant;

namespace ABC150E
{
    class Program
    {
        static void Main(string[] args)
        {
            var N = ReadInt;
            var C = ReadInts;
            C = C.OrderBy(x => -x).ToArray();
            int mod = 1000000007;

            long ans = 0;
            dp = new long[N + 1].Select(x => -1L).ToArray();
            //高速
            for (int i = 0; i < N; i++)
            {
                long cand = C[i] * ModPow(2, N - i - 1, mod) % mod;
                cand *= (ModPow(2, i, mod) + i * ModPow(2, i - 1, mod) % mod) % mod;
                cand %= mod;
                ans = (ans + cand) % mod;
            }
            ans = (ans * ModPow(2, N, mod)) % mod;

            //O(n^2)で確認
            //long anstest = 0;
            //for (int i = 0; i < N; i++)
            //{
            //    for (int k = 0; k <= i; k++)
            //    {
            //        long cand = (C[i] * ModPow(2, N - i - 1, mod)) % mod;
            //        cand = (cand * (k + 1) % mod) * nCk(i, k, mod) % mod;
            //        anstest = (anstest + cand) % mod;
            //    }
            //}
            //anstest = (anstest * ModPow(2, N, mod)) % mod;

            Console.WriteLine(ans);
            //Console.WriteLine(anstest);
        }

        static long[] dp;
        /// <summary>
        /// a^n (mod p)
        /// nが非常に大きい時に有効
        /// </summary>
        /// <param name="a"></param>
        /// <param name="n"></param>
        /// <param name="mod"></param>
        /// <returns></returns>
        public static long ModPow(long a, long n, long p)
        {
            if (n < 0) return 0;
            if (dp[n] != -1) return dp[n];
            if (n == 0) return 1;
            if (n == 1) return a % p;
            if (n % 2 == 1) return (a * ModPow(a, n - 1, p)) % p;

            long t = ModPow(a, n / 2, p);
            return dp[n] = (t * t) % p;
        }

        /// <summary>
        /// 組み合わせ個数
        /// </summary>
        /// <param name="n"></param>
        /// <param name="k"></param>
        /// <param name="p"></param>
        /// <returns></returns>
        public static long nCk(long n, long k, long p)
        {
            long ans = 1;
            k = Math.Min(k, n - k);
            for (long i = n; i > n - k; i--)
            {
                ans = (ans * i) % p;
            }

            for (long i = 1; i <= k; i++)
            {
                ans = (ans * ModPow(i, p - 2, p)) % p;
            }

            return ans;
        }
    }


}

namespace Assistant
{
    static class Input
    {
        static List<string> line = new List<string>();
        static int index = 0;
        static String ReadNext()
        {
            if (line.Count <= index) line.AddRange(Console.ReadLine().Split());
            return line[index++];
        }
        public static int ReadInt => int.Parse(ReadNext());
        public static long ReadLong => long.Parse(ReadNext());
        public static int[] ReadInts => Console.ReadLine().Split().Select(int.Parse).ToArray();
        public static long[] NextLongs => Console.ReadLine().Split().Select(long.Parse).ToArray();
    }

    static class Debug
    {
        //drawmap
        //drawarray
    }
}