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;
#define modulo 1000000007
#define mod(mod_x) ((((long long)mod_x+modulo))%modulo)
#define Inf 1000000000

//aのb乗
int beki(int a,int b){
	int x = 1;
	while(b!=0){
		if(b&1){
			x=mod(x*a);
		}
		a=mod(a*a);
		b>>=1;
	}
	return x;
}

//aの逆元
int gyakugen(int a){
	return beki(a,modulo-2);
}



int main(){
	
	int N;
	cin>>N;
	
	
	vector<int> A(N);
	for(int i=0;i<N;i++)cin>>A[i];
	
	sort(A.rbegin(),A.rend());
	
	int ans = 0;
	
	for(int i=0;i<N;i++){
		int k = mod((i+2) * gyakugen(2));
		k = mod(k * beki(4,N-1));
		k = mod(k * A[i]);
		ans = mod(ans + k);
	}
	
	cout<<mod(ans*2)<<endl;
	
	//cout<<ans<<endl;
	
	
    return 0;
}
