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