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 {
    
    public static List<Integer> P;
    public static void main(String args[]){
        Scanner scanner = new Scanner(System.in);
        final int N = scanner.nextInt();
        final ArrayList<Integer> C = new ArrayList<>();
        final int mod = 1000000007;
        long[] two = new long[N+1];
        two[0] = 1;
        for(int i=0; i<N; i++){
            two[i+1] = (two[i]*2) % mod;
        }
        for(int i=0; i<N; i++){
            C.add(scanner.nextInt());
        }
        Collections.sort(C);
        long ans = 0;
        for(int i=0; i<N; i++){
            long sum;
            if(N-i-1 !=0){
                sum = (two[N-i-2] * (N-i-1)) % mod;
            }else{
                sum = 0;
            }
            sum = (two[N-i-1] + sum) % mod; 
            sum = (two[i] * sum) % mod; 
            sum = (C.get(i) * sum) % mod;
            ans = (ans + sum) % mod;
        }
        ans = (ans * two[N]) % mod;
        System.out.println(ans);
    }
}