SPOJ.com - Problem BIT2

BIT2 - Search Bit Sum

Deepan is a palantir incredible guy who once had a trip in Switzerland. He saw a weird shop where he liked a teddy. But he didn't have enough dollars with him to buy this teddy. But there was a scheme ongoing like solve a equation and get the teddy and pay whatever we wish. That's why this shop is weirdo. He had no eager to buy this teddy as not many people can solve the equation, so he was sure he can get this teddy as soon as he solve the problem. He need your help to solve the equation. You will be given a number N, and you have to tell whether there exist a number K so that sum of BITS in all the numbers from 1 to K is equal to N. If there exist a K, then print it, else print "Does Not Exist." without quotes.

Constraints

1 ≤ N ≤ 10¹⁵, 1 ≤ T ≤ 10⁵.

Input

First line contains t, number of test cases. For each test case, first line contains N.

Output

Output as described above.

Example

Input:
6
282657
377636
472615
567594
662573
1992279

Output:
38050
49299
Does Not Exist.
Does Not Exist.
82953
227394

ID	RESULT	TIME

code...

Submit solution!

Added by:	Rishav Goyal
Date:	2014-02-07
Time limit:	1.223s-2.446s
Source limit:	50000B
Memory limit:	1536MB
Cluster:	Cube (Intel G860)
Languages:	All
Resource:	own

hide comments

2016-10-30 17:22:36
Nice problem!

2016-05-14 16:12:28 rainy jain
Is there any formula for log(n) approach?

Last edit: 2016-05-15 09:45:47

2015-09-19 15:48:45
i ac in log(n)

2014-06-17 04:03:17 Chandan Mittal
learnt just another new thing today
so happy :)

2014-06-17 04:03:17 [Lakshman]
@fitcat Thanks. I checked my algo deeply and find the bug, Now AC with log n.

2014-06-17 04:03:17 newbie
my log(n)^2 giving tle....someone plz help...can log(n)^2 get accepted or i need to find some other algo

Last edit: 2014-02-15 10:03:43

2014-06-17 04:03:17 fitcat
Though I haven't looked into any O(log(n)^2) solutions, I believe and agree that O(log(n)) solution is much simpler. The logic is not hard and in fact quite easy.

@Lakshman: My solution employed unsigned 64 bit integers and there were no overflow problems occurred.

2014-06-17 04:03:17 (Tjandra Satria Gunawan)(æ›¾æ¯…æ˜†)
@Lakshman: maybe my O(log(n)) approach is different than yours..

Last edit: 2014-03-08 12:37:19

2014-06-17 04:03:17 [Lakshman]
@(Tjandra Satria Gunawan)(曾毅昆) some how I am able to write a log n algo but it overflows , and if I use python it gives tle.

2014-06-17 04:03:17 (Tjandra Satria Gunawan)(æ›¾æ¯…æ˜†)
actually O(log(n)) solution is shorter & easier to implement than O(log(n)^2).. the hard part is to get the correct logic..