# The 3n + 1 problem

# The 3n + 1 problem

Consider the following algorithm:

1. input **n**

2. print **n**

3. if **n** = **1** then STOP

4. if **n** is odd then **n** = **3** * **n** + **1**

5. else **n** = **n** / **2**

6. GOTO 2

Given the input **22**, the following sequence of numbers will be printed

**22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1**.

It is conjectured that the algorithm above will terminate (when a **1** is printed) for any integral input value. Despite the simplicity of the algorithm, it is unknown whether this conjecture is true. It has been verified, however, for all integers **n** such that **0** < **n** < **1,000,000** (and, in fact, for many more numbers than this.)

Given an input **n**, it is possible to determine the number of numbers printed (including the **1**). For a given **n** this is called the cycle-length of **n**. In the example above, the cycle length of **22** is **16**.

For any two numbers **i** and **j** you are to determine the maximum cycle length over all numbers between **i** and **j** inclusive.

** Input**

The input will consist of a series of pairs of integers **i** and **j**, one pair of integers per line. All integers will be less than **1,000,000 **and greater than **0**. You can assume that no operation overflows a **32**-bit integer. ** Output**

For each pair of integers **i** and **j** you should output **i** and **j** in the same order in which they appeared in the input. Then print the maximum cycle length for integers between and including **i** and **j**. For each test case print three numbers on one line, separating them by one space.

1 10 100 200 201 210 900 1000

1 10 20 100 200 125 201 210 89 900 1000 174