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;

long long two[400005];
long long mod = 1000000007;
int main(){
	//freopen("i.txt","r",stdin);
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	
	long long n;
	cin >> n;
	
	two[0] = 1;
	for(int i = 1;i < 400000;i++){
		two[i] = two[i-1] * 2LL;
		two[i] %= mod;
	}
	
	long long arr[n];
	for(int i = 0;i < n;i++) cin >> arr[i];
	sort(arr,arr+n);
	reverse(arr,arr+n);
	
	long long base = two[n-1];
	
	long long ans = 0;
	for(int i = 0;i < n;i++){
		ans += base * arr[i];
		ans %= mod;
		if(n >= 2) base += two[n-2];
		base %= mod;
	}
	
	ans *= two[n];
	ans %= mod;
	
	cout << ans;
}
