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 <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <cmath>
#include <utility>
#include <iomanip>
using namespace std;
typedef long long ll;
const ll MOD = 1e9+7;
ll c[200000];
ll beki2[400001];
int main()
{
	int n;
	ll ans=0;
	cin>>n;
	for(int i=0;i<n;i++)
		cin>>c[i];
	sort(c,c+n);
	beki2[0]=1;
	for(int i=1;i<400001;i++)
	{
		beki2[i]=beki2[i-1]*2LL;
		beki2[i]%=MOD;
	}
	for(int i=0;i<n;i++)
	{
		ll tmp=beki2[2*n-2]*(ll)(n-i+1);
		tmp%=MOD;
		//cerr<<tmp<<endl;
		tmp*=c[i];
		tmp%=MOD;
		ans+=tmp;
		ans%=MOD;
	}
	if(ans<0)
		ans+=MOD;
	cout<<ans<<endl;
	return 0;
}