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 <bits/stdc++.h>
#define inf 2147483647
#define MOD 1000000007
using namespace std;
typedef pair<int,int> pii;
int n;
long long arr[202020],ans;
long long g(long long p)
{
	if(p==0) return 1;
	if(p==1) return 2;
	long long ret=g(p/2);
	if(p%2) return ((ret*ret)%MOD*2)%MOD;
	return (ret*ret)%MOD;
}
int main()
{
	scanf("%d",&n);
	for(int i=0;i<n;i++)
	{
		scanf("%lld",arr+i);
	} if(n==1) 
	{
		printf("%lld",(arr[0]*2)%MOD); 
		return 0;
	}
	sort(arr,arr+n); long long m=(long long)n-2;
	long long t=g(m); long long start=(t*(m+3))%MOD;
	for(int i=0;i<n;i++)
	{
		ans=(ans+(start*arr[i])%MOD)%MOD;
		start=(start+MOD-t)%MOD;
	} ans*=g(m+2); ans%=MOD;
	printf("%lld",ans);
}