Odd Digits Reverse Pattern

The program must accept an integer N as the input. The program must print the odd digits from the last digit and remove the last digit of N then again print the odd digits from the last digit and remove the last digit and so on. If there is no odd digit in the original N then print -1 as the output.

Boundary Condition(s):
1 <= N <= 10^7

Input Format:
The first line contains N.

Output Format:
The first line contains either odd digits as per the condition separated by a space or -1.

Example Input/Output 1:
Input:
12345

Output:
5 3 1 3 1 3 1 1 1

Explanation:
In 12345, the odd digits 5,3 and 1 are printed.
After removing the last digit the integer becomes 1234.
In 1234, the odd digits 3 and 1 are printed.
After removing the last digit the integer is 123.
In 123, the odd digits 3 and 1 are printed.
After removing the last digit the integer is 12.
In 12, the odd digit 1 is printed.
After removing the last digit the integer becomes 1.
In 1, the odd digit 1 is printed.

Example Input/Output 2:
Input:
2468

Output:
-1

SOLUTION:-

#include<stdio.h>

#include <stdlib.h>

int main()

{

    

    int n,flag=0;

    scanf("%d",&n);

    

    

    while(n!=0)

    {

        int a=n;

        while(a!=0)

        {

            int x=a%10;

            if(x%2!=0)
         {

            printf("%d ",x);
            flag=1;
           }

            a=a/10;

        }

        n=n/10;

   }

}