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.io.IOException;
import java.io.InputStream;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.NoSuchElementException;

public class Main {

    static final long DIV = 1000000007;
    public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        int n = sc.nextInt();
        long[] c = new long[n];
        for (int i = 0; i < c.length; i++) {
            c[i] = sc.nextLong();
        }

        long[] pow2 = new long[n+1];
        pow2[0] = 1;
        for (int i = 1; i < pow2.length; i++) {
            pow2[i] = (pow2[i-1] * 2) % DIV;
        }

        long ans = 0;
        Arrays.sort(c);
        
        if(c.length == 1){
            ans = 2 * c[0];
            ans %= DIV;
        }else{
            ans += pow2[c.length-1] * c[c.length-1];
            ans %= DIV;
        }

        for (int i = 0; i < c.length - 1; i++) {
            // ans += pow2[i] * c[i] * (n-i-1) * pow2[n-i-2];
            ans += pow2[i] * c[i] * (n-i-1+2) * pow2[n-i-2];
            ans %= DIV;
        }

        ans *= pow2[c.length];
        ans %= DIV;



        System.out.println(ans);


    }



    private static class FastScanner {
        private final InputStream in = System.in;
        private final byte[] buffer = new byte[1024];
        private int ptr = 0;
        private int buflen = 0;

        private boolean hasNextByte() {
            if (ptr < buflen) {
                return true;
            } else {
                ptr = 0;
                try {
                    buflen = in.read(buffer);
                } catch (IOException e) {
                    e.printStackTrace();
                }
                if (buflen <= 0) {
                    return false;
                }
            }
            return true;
        }

        private int readByte() {
            if (hasNextByte())
                return buffer[ptr++];
            else
                return -1;
        }

        private static boolean isPrintableChar(int c) {
            return 33 <= c && c <= 126;
        }

        public boolean hasNext() {
            while (hasNextByte() && !isPrintableChar(buffer[ptr]))
                ptr++;
            return hasNextByte();
        }

        public String next() {
            if (!hasNext())
                throw new NoSuchElementException();
            StringBuilder sb = new StringBuilder();
            int b = readByte();
            while (isPrintableChar(b)) {
                sb.appendCodePoint(b);
                b = readByte();
            }
            return sb.toString();
        }

        public long nextLong() {
            if (!hasNext())
                throw new NoSuchElementException();
            long n = 0;
            boolean minus = false;
            int b = readByte();
            if (b == '-') {
                minus = true;
                b = readByte();
            }
            if (b < '0' || '9' < b) {
                throw new NumberFormatException();
            }
            while (true) {
                if ('0' <= b && b <= '9') {
                    n *= 10;
                    n += b - '0';
                } else if (b == -1 || !isPrintableChar(b)) {
                    return minus ? -n : n;
                } else {
                    throw new NumberFormatException();
                }
                b = readByte();
            }
        }

        public int nextInt() {
            long nl = nextLong();
            if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE)
                throw new NumberFormatException();
            return (int) nl;
        }

        public double nextDouble() {
            return Double.parseDouble(next());
        }
    }

}