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 LL long long
using namespace std;

const int maxn=2e5+7,p=1e9+7;
int c[maxn],n;
LL ans;

LL poww(int x,int y)
{
	LL ans=1,tmp=x;
	while(y) {
		if(y&1) ans=ans*tmp%p;
		tmp=tmp*tmp%p;
		y>>=1;
	}
	return ans;
}

bool cmp(int x,int y)
{
	return x>y;
}

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;++i) scanf("%d",c+i);
	sort(c+1,c+n+1,cmp);
	for(int i=1;i<=n;++i) {
		ans+=1ll*c[i]*(i+1)%p;
		ans%=p;
	}
	ans*=poww(4,n-1);
	ans%=p;
	printf("%lld",ans);
}