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 <iostream>
#include <vector>
#include <bitset>
#include <utility>
#include <string>
#include <queue>
#include <stack>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
 
using namespace std;
typedef long long ll;
 
int main(){
  ll p=1000000007;
  ll N;
  cin >> N;

  vector<int> c(N);
  for(auto &ci:c) cin >> ci;

  sort(c.begin(),c.end());

  //for(auto ci:c) cout << ci << endl;
  
  ll ans=0;
  for(int i=0;i<N;i++){
    //cout << c[i] << " " << N-i+1 << endl;
    ans+=c[i]*(N-i+1);
    ans%=p;
  }

  N=2*N-2;
  ll tmp=2;
  for(int i=0;i<61;i++){
    if(((ll)1<<i)&N){
      ans*=tmp; ans%=p;
    }
    tmp=tmp*tmp; tmp%=p;
  }
    
  cout << ans << endl;
  
  return 0;
}
