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).

Examples

Input

1
1000000000


Output

999999993


Input

2
5 8


Output

124


Input

5
52 67 72 25 79


Output

269312

import java.util.*;
import java.io.*;
import static java.lang.Math.*;

public class Main {

    private static final int MOD = (int)1e9 + 7; //998244353;
    private static final long INF = (long)1e15;

    public Main(FastScanner in, PrintWriter out) {
        int N = in.nextInt();
        long[] C = new long[N];
        for (int i = 0; i < N; i++)
            C[i] = in.nextLong();

        // to get minimum cost, we modify from smallest C[i]
        Arrays.sort(C);

        // for C[i], the first i digit processed has 2^(2i) possibilities
        // for the N - i digits left, the possibility are
        // k out of N - i are different (d[i] must be different):
        // (comb(N - i - 1, k - 1) * 2^k * k) * (2^(N - i - k))
        // the portion in the 1st () is the due to the different digits
        // in the 2nd () is due to the same digits
        // Simplify it, get get: 2^(N - i) * k * comb(N-i-1, k-1)
        // So the total contribution of C[i] is
        // 2^(2i) * Ci * 2^(N-i) * sum(k*comb(N-i-1, k-1) for k in [1, n-i])
        // the sum term is a binomial sum, = 2^(N-i-2)*(n-i+1)
        // so the result is Ci*2^(N-i+2i)*2^(N-i-2)*(N-i+1) = 
        // Ci*2^(2N-2)*(N-i+1)=4^(N-1) * Ci * (N-i+1)
        long p = powm(4, N - 1);
        long res = 0;
        for (int i = 0; i < N; i++) {
            res += C[i] * (N - i + 1) % MOD * p % MOD;
            res %= MOD;
        }
        out.println(res);
    }

   private int gcd(int a, int b) {
        while (true) {
            if (b == 0) return a;
            int tmp = a;
            a = b;
            b = tmp % b;
        }
    }

    private long gcd(long a, long b) {
        while (true) {
            if (b == 0) return a;
            long tmp = a;
            a = b;
            b = tmp % b;
        }
    }

    private long powm(long x, long p) {
        long res = 1;
        while (p > 0) {
            if (p % 2 == 1)
                res = res * x % MOD;
            x = x * x % MOD;
            p /= 2;
        }
        return res;
    }

    public static void main(String[] args) {
        PrintWriter out = new PrintWriter(System.out);
        // Scanner in = new Scanner(
                // new BufferedReader(new InputStreamReader(System.in)));
        FastScanner in = new FastScanner(System.in);

        Main sol = new Main(in, out);
        out.close();
    }
}

// Pair
class Pair<T extends Comparable<? super T>, U extends Comparable<? super U>> implements Comparable<Pair<T, U>> {
    T a;
    U b;
    Pair() { }
    Pair(T a, U b) {
        this.a = a;
        this.b = b;
    }

    @Override
    public String toString() { return "("+a.toString()+", "+b.toString()+")"; }

    @Override
    public int hashCode() { return Objects.hash(a, b); }

    @Override
    public boolean equals(Object obj) {
        if(this == obj) return true;
        if(obj == null) return false;
        if(this.getClass() != obj.getClass()) return false;
        Pair that = (Pair) obj;
        if(this.a.getClass() != that.a.getClass()) return false;
        if(this.b.getClass() != that.b.getClass()) return false;
        if(!this.a.equals(that.a)) return false;
        if(!this.b.equals(that.b)) return false;
        return true;
    }

    @Override
    public int compareTo(Pair<T, U> that) {
        int c = (this.a).compareTo(that.a);
        if(c == 0) c = (this.b).compareTo(that.b);
        return c;
    }
}


class FastScanner{
    private InputStream stream;
    private byte[] buf = new byte[1024];
    private int curChar;
    private int numChars;

    public FastScanner(InputStream stream)
    {
        this.stream = stream;
    }

    int read()
    {
        if (numChars == -1)
            throw new InputMismatchException();
        if (curChar >= numChars){
            curChar = 0;
            try{
                numChars = stream.read(buf);
            } catch (IOException e) {
                throw new InputMismatchException();
            }
            if (numChars <= 0)
                return -1;
        }
        return buf[curChar++];
    }

    boolean isSpaceChar(int c)
    {
        return c==' '||c=='\n'||c=='\r'||c=='\t'||c==-1;
    }

    boolean isEndline(int c)
    {
        return c=='\n'||c=='\r'||c==-1;
    }

    int nextInt()
    {
        return Integer.parseInt(next());
    }

    long nextLong()
    {
        return Long.parseLong(next());
    }

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

    String next(){
        int c = read();
        while (isSpaceChar(c))
            c = read();
        StringBuilder res = new StringBuilder();
        do{
            res.appendCodePoint(c);
            c = read();
        }while(!isSpaceChar(c));
        return res.toString();
    }

    String nextLine(){
        int c = read();
        while (isEndline(c))
            c = read();
        StringBuilder res = new StringBuilder();
        do{
            res.appendCodePoint(c);
            c = read();
        }while(!isEndline(c));
        return res.toString();
    }
}
