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;

int main() {
  int64_t N,Q=1000000007,Z=0;
  cin>>N;
  vector<int64_t> p(N);
for (int i=0;i<N;i++) {
  cin>>p.at(i);
}
  sort(p.begin(), p.end());
  int64_t i=2;
  int64_t M=(N-1)*2; //ここから
       int64_t a=1,b=0,c=1;
       while(M>0){
         a*=2;
         if(M%a!=0){
           M-=a/2;
           int64_t d=i;
           for(int j=0;j<b;j++){
            d=d*d;
             d=d%Q;
          }
           c=c*d;
          c=c%Q;
         }
        b++;
      } //ここまで
  
    for(int i=0;i<N;i++){
      Z=(Z+(((p.at(i)*(N+1-i))%Q)*c)%Q)%Q;
    }
    cout<<Z<<endl;
  
}