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 <algorithm>
#define rep(i,n) for(int i=0; i<(int)(n); i++)

using namespace std;
typedef long long LL;
const LL mod=1e9+7;

LL modpow(LL x, LL n){
	LL r=1;
	while(n){
		if(n&1) r=r*x%mod;
		x=x*x%mod;
		n>>=1;
	}
	return r;
}

int main(){
  int N;
  cin >> N;
  vector<LL> C(N);
  rep(i,N) cin >> C[i];
  sort(C.begin(),C.end());
  vector<LL> P(N);
  rep(i,N){
    P[i]=modpow(2,2*N-2)*(N+1-i);
    P[i]%=mod;
  }
  LL ans=0;
  rep(i,N){
    ans+=C[i]*P[i];
    ans%=mod;
  }
  cout << ans << endl;

  return 0;
}
