### PROBLEM LINK:

### DIFFICULTY:

CAKEWALK

### PREREQUISITES:

None.

### EXPLANATION

This problem can be solved by Brute Force.

Dividing the number by 2 in a loop and checking for remainder would suffice.

But the trick is with smaller source code.A common BIT-TRICK of using this expression ((n)&(n-1)) would be handy,

if one wants to shorten the source code.