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 mod_pow(long long a,long long b,long long mod){
	long long res=1,c=a;
	while(b){
		if(b&1) res=res*c%mod;
		c=c*c%mod;
		b/=2;
	}
	return res;
}
int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	
	int n;
	cin >> n;
	vector<long long> a(n);
	for(int i=0;i<n;i++) cin >> a[i];
	long long mod=1000000007;
	sort(a.begin(),a.end());
	long long res=0;
	for(int i=0;i<n;i++){
		//1+binom(n-1,1)*2+binom(n-1,2)*3+
		long long coef=mod_pow(2,n-i-1,mod)*(n-i+1)%mod*((mod+1)/2)%mod;
		coef=coef*mod_pow(2,i,mod)%mod;
		res=(res+coef*a[i]%mod)%mod;
	}
	res=res*mod_pow(2,n,mod)%mod;
	cout << res << "\n";
}