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
n=int(input())
c=sorted(list(map(int,input().split())))
#n=200000
#c=list(range(n))
mod=pow(10,9)+7
# T=000000..とおいて計算し最後に2^nをかける。
# 各コストCiごとにそれが何回足されるかを求める。
# 一番大きいコストc[-1]の係数は必ず1にするのが最適。同様に一番小さいコストc[0]は必要な係数の中で一番大きなものをかけるのがいい。
ans=0
for i,ci in enumerate(c):
  l=i
  r=n-i
  b=pow(2,r-1,mod)+(r-1)*pow(2,r-2,mod) if r>1 else 1
  b*=pow(2,l,mod)
  ans+=ci*b
  ans%=mod
ans*=pow(2,n,mod)
print(ans%mod)
