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 N=2e5+5;
const int MD=1e9+7;
int a[N];
int quick_pow(int x,int y) {
	int ans=1;
	while(y) {
		if(y&1) ans=1LL*ans*x%MD;
		y>>=1;
		x=1LL*x*x%MD;
	}
	return ans;
}
void add(int &x,int y) {
	x+=y;
	if(x>=MD) x-=MD;
}
int main() {
	int n;
	scanf("%d",&n);
	for(int i=0;i<n;i++) {
		scanf("%d",&a[i]);
	}
	if(n==1) printf("%d\n",1LL*2*a[0]%MD);
	else {
		sort(a,a+n);
		int ans=0,x=quick_pow(2,n-2),t=1LL*x*2%MD;
		for(int i=n-1;i>=0;i--) {
			add(ans,1LL*a[i]*t%MD);
			add(t,x);
		}
		printf("%d\n",1LL*ans*quick_pow(2,n)%MD);
	}
	return 0;
}