The 3-Digit Kaprekar Number

Problem Description

A "black hole number" (also called a trap number, or the "Kaprekar problem") is a class of numbers with a peculiar transform property. Any three-digit number whose digits are not all the same will, after a finite number of "rearrange and subtract" operations, always reach $495$. The resulting $495$ is the 3-digit black hole number. The so-called "rearrange and subtract" operation means: form the largest number and the smallest number obtainable by rearranging the digits of the number, and subtract the smallest from the largest. ($6174$ is the 4-digit black hole number.)

For example, for the three-digit number $207$:

1st rearrange-and-subtract: $720 - 27 = 693$; 2nd: $963 - 369 = 594$; 3rd: $954 - 459 = 495$; after that it stays at the black hole number $495$. If the three digits of a three-digit number are all the same, one transform gives $0$.

Given any three-digit number, write a program that outputs the rearrange-and-subtract process.

Input Format

The input gives one three-digit number on one line.

Output Format

Output the rearrange-and-subtract process in the following format:

index: largest number after rearrangement - smallest number after rearrangement = difference

The index starts from $1$ and continues until $495$ appears on the right side of the equals sign.

Sample Input

123

Sample Output

1: 321 - 123 = 198
2: 981 - 189 = 792
3: 972 - 279 = 693
4: 963 - 369 = 594
5: 954 - 459 = 495

Sample Explanation

The input three-digit number is $123$. First, rearrange the digits $1, 2, 3$ into the largest number $321$ and the smallest number $123$, subtract to get $198$. Then repeat the operation on $198$: rearrange into the largest number $981$ and smallest $189$, subtract to get $792$. Continuing, we get $693, 594$, and finally $495$, and the process ends.

Constraints

The input three-digit number satisfies $100 \le n \le 999$.