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 <cmath>
#include <vector>
#include <map>
#include <stack>
#include <queue>
#include <set>
#include <algorithm>
#include <iomanip>
#include <string.h>

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define ALL(a)  (a).begin(),(a).end()

typedef long long lint;

using namespace std;
lint p=1e9+7;

int main(){
  int N;
  cin>>N;
  lint C[N];
  REP(i,N)cin>>C[i];
  sort(C,C+N);
  lint ans=0;
  REP(i,N){
    ans+=C[i]*(lint)(1+N-i);
    ans%=p;
  }
  REP(i,2*N-2){
    ans*=2;
    ans%=p;
  }
  cout<<ans<<endl;
  return 0;
}
