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
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#define pi acos(-1)
#define p 1000000007
using namespace std;
int N,c[201010];
long long f2[303030],ans,F[201010];
int n,m;
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	scanf("%d",&c[i]);
	sort(c+1,c+n+1);
	
	f2[0]=1;
	for(int i=1;i<=n;i++)
	f2[i]=f2[i-1]*2LL%p;
	
	F[0]=1;
	for(int i=1;i<=n;i++) F[i]=(2LL*F[i-1]%p+f2[i-1])%p;
	for(int i=1;i<=n;i++)
		(ans+=1LL*c[i]*f2[i-1]%p*F[n-i]%p)%=p;	
	printf("%lld\n",1LL*ans*f2[n]%p);
	return  0;
}
		