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 ll long long
const int N=200005,Mod=1e9+7;
int a[N];
inline int qpow(int a,int b){
	int ans=1;
	while (b){
		if (b&1) ans=1ll*ans*a%Mod;
		a=1ll*a*a%Mod,b>>=1;
	}
	return ans;
}
int main (){
	int n;scanf ("%d",&n);
	for (int i=1;i<=n;i++) scanf ("%d",&a[i]);
	sort(a+1,a+n+1);int ans=0,all=1ll*qpow(2,n)*qpow(2,n)%Mod;
	int inv2=qpow(2,Mod-2);all=1ll*all*inv2%Mod;
	for (int i=1;i<=n;i++)
		ans=(ans+1ll*all*(n-i+2)%Mod*inv2%Mod*a[i])%Mod;
	printf ("%d",ans);
	return 0;
}