Let's call the following process a transformation of a sequence of length n.

If the sequence is empty, the process ends. Otherwise, append the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor) (GCD) of all the elements of the sequence to the result and remove one arbitrary element from the sequence. Thus, when the process ends, we have a sequence of n integers: the greatest common divisors of all the elements in the sequence before each deletion.

You are given an integer sequence 1, 2, ..., n. Find the lexicographically maximum result of its transformation.

A sequence a_1, a_2, …, a_n is lexicographically larger than a sequence b_1, b_2, …, b_n, if there is an index i such that a_j = b_j for all j < i, and a_i > b_i.

Input

The first and only line of input contains one integer n (1≤ n≤ 10^6).

Output

Output n integers — the lexicographically maximum result of the transformation.

Examples

Input

3


Output

1 1 3 

Input

2


Output

1 2 

Input

1


Output

1 

Note

In the first sample the answer may be achieved this way:

  * Append GCD(1, 2, 3) = 1, remove 2. 
  * Append GCD(1, 3) = 1, remove 1. 
  * Append GCD(3) = 3, remove 3. 



We get the sequence [1, 1, 3] as the result.
from __future__ import division, print_function
# import threading
# threading.stack_size(2**27)
# import sys
# sys.setrecursionlimit(10**7)
from sys import stdin, stdout
import bisect            #c++ upperbound
import math
import heapq
def modinv(n,p):
    return pow(n,p-2,p)
def cin():
    return map(int,sin().split())
def ain():                           #takes array as input
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input())
import math  
def Divisors(n) : 
    l = []  
    for i in range(1, int(math.sqrt(n) + 1)) :
        if (n % i == 0) : 
            if (n // i == i) : 
                l.append(i) 
            else : 
                l.append(i)
                l.append(n//i)
    return l
p={}
q=[]
def dfs(n,d,v):
    global p
    global q
    v[n]=1
    x=d[n]
    q.append(n)
    j=0
    for i in x:
        if i not in v:
            f=dfs(i,d,v)
            p[i]=f
            j+=f
    return j+1

"""*******************************************************"""
def main():
    n=inin()
    a=[]
    x=1
    b=[1,1,3]
    while(n>0):
        if(n==3):
            for i in b:
                a.append(i*x)
            break
        for i in range(int(math.ceil(n/2))):
            a.append(x)
        n//=2
        x*=2
    print(*a)



######## Python 2 and 3 footer by Pajenegod and c1729
 
# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.
 
# So on cf, use PyPy2 for best string performance.
 
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange
 
import os, sys
from io import IOBase, BytesIO
 
BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
 
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None
 
    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s
 
    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()
 
    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            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')
 
# Cout implemented in Python
import sys
class ostream:
    def __lshift__(self,a):
        sys.stdout.write(str(a))
        return self
cout = ostream()
endl = '\n'
 
# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero = 0):
    conv = ord if py2 else lambda x:x
    A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'0' [0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-' [0]: sign = -1
            elif s[i] != b'\r' [0]:
                A.append(sign*numb)
                numb = zero; sign = 1
            i += 1
    except:pass
    if s and s[-1] >= b'0' [0]:
        A.append(sign*numb)
    return A

# threading.Thread(target=main).start()
if __name__== "__main__":
  main()