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>

using namespace std ;

const long long mod = 1e9 + 7ll ;
const int MAX = 2e5 + 10 ;

long long arr[MAX] , P[MAX] ;
long long n ;

int main()
{
	ios_base::sync_with_stdio(0) ;
	cin.tie(0) ;
	P[0] = 1ll ;
	for(int i = 1 ; i < MAX ; ++i)
		P[i] = (P[i-1] * 2ll) % mod ;
	cin>>n ;
	for(int i = 0 ; i < n ; ++i)
		cin>>arr[i] ;
	sort(arr , arr + n) ;
	long long ans = 0ll ;
	for(int i = 0 ; i < n ; ++i)
	{
		long long a = n-i-1 ;
		long long x = (a * P[a-1]) % mod ;
		x = (x + P[a]) % mod ;
		x = (x * P[a+1]) % mod ;
		x = (x * P[i]) % mod ;
		x = (x * P[i]) % mod ;
		long long y = (arr[i] * x) % mod ;
		ans = (ans + y) % mod ;
	}
	return cout<<ans<<"\n" , 0 ;
}		