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>
#define M 1000000007LL
using namespace std;
typedef long long ll;
ll pot(ll a,ll b){
	if(b == 0)
		return 1;
	if(b&1)
		return (a*pot((a*a)%M,b/2))%M;
	return pot((a*a)%M,b/2)%M;
}
int main(){
	ll n;
	scanf("%lld",&n);
	if(n == 1){
		scanf("%lld",&n);
		n = (n*2LL)%M;
		printf("%lld\n",n);
	}
	else{
		ll res = 0;
		ll ini = pot(2,n-1);
		ll paso = pot(2,n-2);
		vector<ll> val(n);
		for(int i = 0; i < n; i++)
			scanf("%lld",&val[i]);
		sort(val.rbegin(), val.rend());
		for(int i = 0; i < n; i++){
			res = (res + (val[i]*ini)%M)%M;
			ini = (ini + paso)%M;
		}
		res = (res * pot(2,n))%M;
		printf("%lld\n", res);
	}
	return 0;
}