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
// I SELL YOU...! 
#include<iostream>
#include<vector>
#include<algorithm>
#include<functional>
#include<queue>
#include<chrono>
#include<iomanip>
#define MOD 1000000007
using namespace std;
using ll = long long;
using P = pair<ll,ll>;
void init_io(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << setprecision(10);
}
signed main(){
  init_io();
  ll n,ans=1,tp=0;
  cin >> n;
  vector<ll> c(n);
  for(int i=0;i<n;i++){
    cin >> c[i];
    if(i!=0){
      ans *= 4;
      ans %= MOD;
    }
  }
  sort(c.begin(),c.end(),greater<ll>());
  for(int i=0;i<n;i++){
    tp += ((i+2)*(c[i]))%MOD;
    tp%=MOD;
  }
  cout << (ans*tp)%MOD<<endl;
}
