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 int M = 1e9+7;
long long pwr( long long a , long long b , long long m = M ){
	
	long long ret = 1;
	a %= m , b %= m;

	while( b>0 ){
		if(b&1)	ret = ret*a % m;
		a = a*a % m;
		b >>= 1;
	}

	return ret;
}

int main(){

	ios_base::sync_with_stdio(0);
	cout.tie(0);
	cin.tie(0);

	int n;
	cin>>n;

	vector<int> a(n);
	for(int i=0 ; i<n ; ++i)
		cin>>a[i];

	sort(a.begin(),a.end(),greater<int>());
	long long ans = 0;

	for(int i=0 ; i<n ; ++i)
		ans = ( ans + 1ll*(i+2)*a[i] ) % M;

	ans = ans * pwr(2,2*n-2) % M;
	cout<< ans ;
}