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.*;

public class Main {

    static int mod = 1_000_000_007;
    static long pow(long a,int n) {
        long res=1;
        while (n>0) {
            if ((n-n/2*2)==1) {
                res=res*a%mod;
            }
            a=a*a%mod;
            n>>=1;
        }
        return res;
    }

    static long modinv(long n) {
        return pow(n, mod-2);
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = Integer.parseInt(sc.next());
        long[] C = new long[N];
        for (int i=0;i<N;i++) {
            C[i] = Integer.parseInt(sc.next());
        }
        long[] val = new long[N];
        long a = pow(2, N-1);
        long b = pow(2, N-2);
        for (int i=0;i<N;i++) {
            val[i] = (a+b*i%mod)%mod;
        }
        Arrays.sort(C);
        long ans = 0L;
        for (int i=0;i<N;i++) {
            ans = (ans+C[i]*val[N-1-i]%mod)%mod;
        }
        System.out.println(ans*pow(2, N)%mod);
    }
}