Problem:
Score : 500 points

Problem Statement

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).

Sample Input 1

1
1000000000

Sample Output 1

999999993

There are two pairs (S, T) of different sequences of length 2 consisting of 0 and 1, as follows:

S = (0), T = (1): by changing S_1 to 1, we can have S = T at the cost of 1000000000, so f(S, T) = 1000000000.

S = (1), T = (0): by changing S_1 to 0, we can have S = T at the cost of 1000000000, so f(S, T) = 1000000000.

The sum of these is 2000000000, and we should print it modulo (10^9+7), that is, 999999993.

Sample Input 2

2
5 8

Sample Output 2

124

There are 12 pairs (S, T) of different sequences of length 3 consisting of 0 and 1, which include:

S = (0, 1), T = (1, 0)

In this case, if we first change S_1 to 1 then change S_2 to 0, the total cost is 5 \times 2 + 8 = 18. We cannot have S = T at a smaller cost, so f(S, T) = 18.

Sample Input 3

5
52 67 72 25 79

Sample Output 3

269312

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#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);
	for(int i=0;i<n;i++)
	{
		ans=(ans+start*arr[i])%MOD;
		start-=t;
	} ans*=g(m+2); ans%=MOD;
	printf("%lld",ans);
}