human
Can you provide a Python program to determine how many teams he should invite so that the tournament needs exactly n games given the following:
Daniel is organizing a football tournament. He has come up with the following tournament format: 

  1. In the first several (possibly zero) stages, while the number of teams is even, they split in pairs and play one game for each pair. At each stage the loser of each pair is eliminated (there are no draws). Such stages are held while the number of teams is even. 
  2. Eventually there will be an odd number of teams remaining. If there is one team remaining, it will be declared the winner, and the tournament ends. Otherwise each of the remaining teams will play with each other remaining team once in round robin tournament (if there are x teams, there will be  games), and the tournament ends. 


For example, if there were 20 teams initially, they would begin by playing 10 games. So, 10 teams would be eliminated, and the remaining 10 would play 5 games. Then the remaining 5 teams would play 10 games in a round robin tournament. In total there would be 10+5+10=25 games.

Daniel has already booked the stadium for n games. Help him to determine how many teams he should invite so that the tournament needs exactly n games. You should print all possible numbers of teams that will yield exactly n games in ascending order, or -1 if there are no such numbers.

Input

The first line contains a single integer n (1 ≤ n ≤ 1018), the number of games that should be played.

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.

Output

Print all possible numbers of invited teams in ascending order, one per line. If exactly n games cannot be played, output one number: -1.

Examples

Input

3

Output

3
4

Input

25

Output

20

Input

2

Output

-1
Solve the task in PYTHON.
assistant
#Here is Python program to determine how many teams he should invite so that the tournament needs exactly n games.

# -*- coding: utf-8 -*-
"""
Spyder Editor

This temporary script file is located here:
/home/zangetsu/.spyder2/.temp.py
"""

from __future__ import division;
from bisect import *;
import sys;
from math import *;
from fractions import *;
from itertools import *; 
import io;
import re;

INF = 987654321987654321987654321;

def readint(delimiter=' ') :
    return map(int, raw_input().split(delimiter));

def readstr(delimiter=' ') :
    return raw_input().split(delimiter);

def readfloat(delimiter=' ') :
    return map(float, raw_input().split(delimiter));

def index(a, x):
    'Locate the leftmost value exactly equal to x'
    i = bisect_left(a, x)
    if i != len(a) and a[i] == x:
        return i
    raise ValueError

def find_lt(a, x):
    'Find rightmost value less than x'
    i = bisect_left(a, x)
    if i:
        return a[i-1]
    raise ValueError

def find_le(a, x):
    'Find rightmost value less than or equal to x'
    i = bisect_right(a, x)
    if i:
        return a[i-1]
    raise ValueError

def find_gt(a, x):
    'Find leftmost value greater than x'
    i = bisect_right(a, x)
    if i != len(a):
        return a[i]
    raise ValueError

def find_ge(a, x):
    'Find leftmost item greater than or equal to x'
    i = bisect_left(a, x)
    if i != len(a):
        return a[i]
    raise ValueError

def bin_search(a, x, left, right) :

    while left<=right :
        mid = (left + right)//2;
        
        if a[mid] == x :
            return mid;
        elif a[mid] < x :
            left = mid + 1; 
        elif a[mid] > x :
            right = mid - 1;
                       
        pass
    
    return -1;
    pass

def printf(format, *args):
    """Format args with the first argument as format string, and write.
    Return the last arg, or format itself if there are no args."""
    sys.stdout.write(str(format) % args)
    pass

if __name__ == '__main__':

    n = readint()[0];
    #n = float(n);
    limit = int(log(n+1, 2));
    
    def f(x, a) :
        return (x)*(x-1)//2 + x*(2**a-1);
        
  ...