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.math.*;
public class Main{
	public static void main(String[] args){
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		long mod=1000000007;
		long[] c=new long[n];
		for(int i=0; i<n; i++){
			c[i]=sc.nextLong();
		}
		long two=1;
		long[] mm=new long[n+1];
		for(int i=0; i<=n; i++){
			mm[i]=two;
			two*=2;
			two%=mod;
		}
		Arrays.sort(c);
		long ans=(c[n-1]*mm[n-1])%mod;
		for(int i=0; i<n-1; i++){
			ans+=c[i]*(((mm[n-1-i]+(mm[n-2-i]*(n-1-i))%mod)*mm[i])%mod);
			ans%=mod;
		}
		ans*=mm[n];
		System.out.println(ans%mod);
	}
}
