313_C. Ilya and Matrix
Ilya is a very good-natured lion. He likes maths. Of all mathematical objects, his favourite one is matrices. Now he's faced a complicated matrix problem he needs to solve.

He's got a square 2n × 2n-sized matrix and 4n integers. You need to arrange all these numbers in the matrix (put each number in a single individual cell) so that the beauty of the resulting matrix with numbers is maximum.

The beauty of a 2n × 2n-sized matrix is an integer, obtained by the following algorithm:

  1. Find the maximum element in the matrix. Let's denote it as m. 
  2. If n = 0, then the beauty of the matrix equals m. Otherwise, a matrix can be split into 4 non-intersecting 2n - 1 × 2n - 1-sized submatrices, then the beauty of the matrix equals the sum of number m and other four beauties of the described submatrices. 



As you can see, the algorithm is recursive.

Help Ilya, solve the problem and print the resulting maximum beauty of the matrix.

Input

The first line contains integer 4n (1 ≤ 4n ≤ 2·106). The next line contains 4n integers ai (1 ≤ ai ≤ 109) — the numbers you need to arrange in the 2n × 2n-sized matrix.

Output

On a single line print the maximum value of the beauty of the described matrix.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Examples

Input

1
13


Output

13


Input

4
1 2 3 4


Output

14

Note

Consider the second sample. You need to arrange the numbers in the matrix as follows:
    
    
      
    1 2  
    3 4  
    

Then the beauty of the matrix will equal: 4 + 1 + 2 + 3 + 4 = 14.
```python
import os
import sys
from io import BytesIO, IOBase
 
BUFSIZE = 8192
 
 
class FastIO(IOBase):
    newlines = 0
 
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
 
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
 
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)
 
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
################################
n=int(input())
from math import log
l=list(map(int,input().split()))
l.sort(reverse=True)
a=int(log(n,4))
if n==1:
    print(l[0])
else:
    ans=l[0]*(a+1)
    i=1
    c=0
    while i<n:
        if a>1:
            for j in range((4**c)*3):
                ans+=l[i+j]*a
            a-=1
            i+=(4**c)*3
            c+=1
        else:
            ans+=l[i]
            i+=1
    print(ans)

```