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 n,p,total;
long long arr[200001];
long long power(long long x,long long y){
    long long ans=1;
    x%=p;
    while (y>0){
        if (y%2==1) ans=(ans*x)%p;
        y/=2;
        x=(x*x)%p;
    }
    return ans;
}
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin>>n;
    for (int i=0;i<n;i++) cin>>arr[i];
    p=1e9+7;
    sort(arr,arr+n);
    reverse(arr,arr+n);
    long long x=power(2,n*2-1);
    long long y=x;
    y*=500000004;
    y%=p;
    for (int i=0;i<n;i++){
        total+=arr[i]*x;
        total%=p;
        x+=y;
        x%=p;
    }
    cout<<total;
}