Tag(s):

## Basic Programming, Bit manipulation, Easy

Problem
Editorial
Analytics

The problem is straight and simple.
Given a number $X$ ,find how many positive $A$ ( $A\gt 0)$ exists, such that
1. $A$ $\newcommand*\xor{\mathbin{\oplus}}$ $X$ =$A$ + $X$
2. $A$ $\lt$ $X$
Input:
The first line of the input will contain $T$ , the number of test-cases.
Next $T$ lines will contain integer $X$ .
Output:
For each test-case , output the answer in a separate line.

Constraints:

• $1 \le T \le 10^5$
• $1 \le X \le 10^7$
Note: $\newcommand*\xor{\mathbin{\oplus}}$ is the bitwise Exclusive-OR operator( XOR ). and + is the usual addition symbol.

SAMPLE INPUT
4
1
2
3
4

SAMPLE OUTPUT
0
1
0
3

Explanation

Explanation for 1st test case:
For $1$ there is no such number.
For 2nd case:
For $2$ we have only one such number $1$ $\newcommand*\xor{\mathbin{\oplus}}$ $2$ = $3$
For 3rd case:
For number $3$ we have no such number .
For 4th case:
For number $4$ we have $3$ $\oplus$ $4$ = $7$ , $2$ $\oplus$ $4$ = $6$ and $1$ $\oplus$ $4$ = $5$. Hence the answer is 3.

Time Limit: 1.0 sec(s) for each input file.
Memory Limit: 256 MB
Source Limit: 1024 KB
Marking Scheme: Marks are awarded when all the testcases pass.
Allowed Languages: C, C++, Clojure, C#, D, Erlang, F#, Go, Groovy, Haskell, Java, Java 8, JavaScript(Rhino), JavaScript(Node.js), Lisp, Lisp (SBCL), Lua, Objective-C, OCaml, Octave, Pascal, Perl, PHP, Python, Python 3, R(RScript), Racket, Ruby, Rust, Scala, Scala 2.11.8, Swift, Visual Basic

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