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;
using ll=long long;
using Graph=vector<vector<int>>;
#define MAX 5100000
#define MOD 1000000007
#define INF 100000000000000

int main(){
  ll N;
  cin>>N;
  vector<pair<ll,int>> C(N);
  for(int i=0;i<N;i++){
    cin>>C.at(i).first;
    C.at(i).second=i;
  }
  sort(C.begin(),C.end());
  ll ans=0;
  vector<ll> res(N+2,1);
  for(int i=0;i<N+1;i++){
    res.at(i+1)=2*res.at(i)%MOD;
  }
  ll pat=res.at(N);
  for(ll i=0;i<N-1;i++){
    ans+=(C.at(i).first*pat%MOD)*((((N-i-1)*res.at(N-i-2)+res.at(N-i-1))%MOD)*res.at(i)%MOD)%MOD;
    ans%=MOD;
  }
  ans+=(C.at(N-1).first*pat%MOD)*res.at(N-1)%MOD;
  ans%=MOD;
  cout<<ans<<endl;
}
