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;
typedef long long ll;
#define rep(i,a) for(int i=0;i<(a);i++)
const ll MOD=1000000007;
//const ll MOD=998244353;

ll fac[202020];

int main(){
  
  fac[0]=1;
  rep(i,200002) fac[i+1]=fac[i]*2%MOD;
  
  int N; cin>>N;
  ll C[N]; rep(i,N) cin>>C[i];
  sort(C,C+N);
  reverse(C,C+N);
  
  ll ans=0;
  rep(i,N){
    ll mul=fac[i];
    if(i) mul=(mul+i*fac[i-1])%MOD;
    mul=(mul*C[i])%MOD*fac[N-1-i]%MOD;
    ans=(ans+mul)%MOD;
  }
  cout<<ans*fac[N]%MOD<<endl;
  return 0;
}