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;
static const int64_t mod=1000000007;
int64_t f(int64_t x,int64_t y){
if(y==0)
  return 1;
  else{
  if(y%2==0)
    return (f(x,y/2)*f(x,y/2))%mod;
    else
      return (x*((f(x,y/2)*f(x,y/2))%mod))%mod;
  }
}
int main(){
int64_t N;
  cin>>N;
  vector<int64_t>C(N);
  for(int64_t i=0;i<N;i++)
    cin>>C.at(i);
  sort(C.begin(),C.end());
  if(N==1)
    cout<<(2*C.at(0))%mod<<endl;
  else{
  int64_t ans=0;
    int64_t a=f(2,N-1);
    int64_t b=f(2,N-2);
    for(int64_t i=0;i<N;i++){
    ans=(ans+(((N-1-i)*b+a)%mod)*C.at(i))%mod;
    }int64_t c=f(2,N);
    ans=(ans*c)%mod;
    cout<<ans<<endl;
 }
   return 0;
}