MS takes several forms, with new symptoms either happening in isolated attacks (relapsing forms) or constructing up over time (progressive forms).Between attacks, symptoms may disappear completely; however, permanent neurological problems often remain, particularly with the advancement of the disease.

COMING SOON!

```
/**
* Check if a given number is multiple of 3
* Basic way of doing is : if sum of digits of number is multiple of 3,
* number is divisible by 3.
*
* However another efficient way of doing it is:
* Count the number of set bits at even positions.
* Count the number of set bits at odd positions.
* if difference is multiple of 3, number will be multiple of 3.
*
* Proof:
* We can prove it by taking the example of 11 in decimal numbers.
* (It will apply to 3 in binary numbers as well.)
* AB = 10A + B ( A and B are digits of number AB)
* AB = 11A + (B - A)
* Clearly if (B-A) is multiple of 11, number would be muliple of 11.
* We can do it similarly for 3 digits
* ABC = 100A + 10B + C
* = 99A + A + 11B - B + C
* = 11(9A + B) + (A - B + C)
* Clearly if A-B+C is multiple of 11, ABC would be multiple as well.
* similarly for 4 digits
* ABCD = 1000A + 100B + 10C + D
* = (1001A – 999B + 11C) + (D + B – A -C )
* So, if (B + D – A – C) is a multiple of 11 then is ABCD.
*/
#include <iostream>
bool is_multiple_of_3( int n )
{
// change to positive if negative
if ( n < 0 ) {
n = -n;
}
if ( n == 0 ) {
return true;
}
if ( n == 1 ) {
return false;
}
int even_count = 0;
int odd_count = 0;
while ( n ) {
if ( n & 1 ) {
++odd_count;
}
n >>= 1;
if ( n & 1 ) {
++even_count;
}
n >>= 1;
}
return is_multiple_of_3( even_count - odd_count );
}
int main()
{
int num;
std::cout << "Enter a number:";
std::cin >> num;
if (is_multiple_of_3(num)) {
std::cout << num << " is multiple of 3" << std::endl;
} else {
std::cout << num << " is not a multiple of 3" << std::endl;
}
return 0;
}
```