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.lang.*;

class Mod{
    static final long MOD = 1000_000_007;
    static final long[] fact = makeFactorialArray(1000000);
    static final long[] factInv = makeInversedArray(fact);

    public static long add(long a, long b){
        long tmp = (a+b)%MOD;
        return tmp>=0 ? tmp : tmp+MOD;
    }
    public static long adds(long... ar){
        long ans=0;
        for(long x:ar) ans = Mod.add(ans, x);
        return ans;
    }
    public static long sub(long a, long b){
        return add(a,-b);
    }
    public static long mult(long a, long b){
        long tmp = (a*b)%MOD;
        return tmp>=0 ? tmp : tmp+MOD;
    }
    public static long mults(long... ar){
        long ans=1;
        for(long x:ar) ans = Mod.mult(ans, x);
        return ans;
    }
    public static long power(long a, long x){ //calculate a^x
        if(x<0)return 0;
        if(x==0)return 1;
        if(x%2==0) {
            long half = power(a,x/2);
            return mult(half,half);
        }
        return (a*power(a,x-1)) % MOD;
    }
    public static long inverse(long a){
        return power(a,MOD-2);
    }
    public static long div(long a, long b){
        return mult(a, inverse(b));
    }
    public static long[] makeFactorialArray(int size){
        long[] array = new long[size];
        array[0]=1;
        for(int i=1;i<size;i++){
            array[i]=mult(array[i-1],i);
        }
        return array;
    }
    public static long[] makeInversedArray(long[] original){
        long[] array = new long[original.length];
        for(int i=0;i<original.length;i++){
            array[i] = inverse(original[i]);
        }
        return array;
    }
    public static long combination(long n, long r){
        int N = (int)n,  R = (int)r;
        if(n<0 || r<0 || n<r) return 0;
        return mult(fact[N], mult(factInv[R],factInv[N-R]));
    }
}
public class Main {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        long[] C = new long[N];
        for(int n=0; n<N; n++) C[n]=sc.nextLong();
        Arrays.sort(C);

        long ans = 0;
        for(int n=0; n<N; n++){
            ans = Mod.add(ans,
                          Mod.mults(N-1-n==0 ? 1L : (Mod.mults(Mod.power(2,N-2-n), N+1-n)),
                                    Mod.power(2,n),
                                    C[n]));
        }
        System.out.println(Mod.mult(ans, Mod.power(2, N)));
    }

}
