3918：A to Z Numerals

Roman numerals use symbols I, V, X, L, C, D, and M with values 1, 5, 10, 50, 100, 500, and 1000 respectively. There is an easy evaluation rule for them:
Rule ∆: Add together the values for each symbol that is either the rightmost or has a symbol of no greater value directly to its right. Subtract the values of all the other symbols.
For example: MMCDLXIX = 1000 + 1000 - 100 + 500 + 50 + 10 - 1 + 10 = 2469.
Further rules are needed to uniquely specify a Roman numeral corresponding to a positive integer less than 4000:
1. The numeral has as few characters as possible. (IV not IIII)
2. All the symbols that make positive contributions form a non-increasing subsequence. (XIV, not VIX)
3. All subtracted symbols appear as far to the right as possible. (MMCDLXIX not MCMDLIXX)
4. Subtracted symbols are always for a power of 10, and always appear directly to the left of a symbol 5 or 10 times as large that is added. No subtracted symbol can appear more than once in a numeral.
Rule 4 can be removed to allow shorter numerals, and still use the same evaluation rule: IM = -1 + 1000 = 999, ILIL = -1 + 100 + -1 + 100 = 198, IVL = -1 -5 + 100 = 94. This would not make the numerals unique, however. Two choices for 297 would be CCVCII and ICICIC. To eliminate the second choice in this example, Rule 4 can be replaced by
4'. With a choice of numeral representations of the same length, use one with the fewest subtracted symbols.
Finally, replace the Roman numeral symbols to make a system that is more regular and allows larger numbers: Assign the English letter symbols a, A, b, B, c, C, …, y, Y, z, and Z to values 1, 5, 10, 5(10), 10², (5)10², ..., 10²⁴, (5)10²⁴, 10²⁵, and (5)10²⁵ respectively. Though using the whole alphabet makes logical sense, your problem will use only symbols a-R for easier machine calculations. (R = (5)10¹⁷.)
With the new symbols a-Z, the original formation rules 1-3, the alternate rule 4', and the evaluation rule ∆, numerals can be created, called A to Z numerals. Examples: ad = -1 + 1000 = 999; aAc = -1 - 5 + 100 = 94.

The input starts with a sequence of one or more positive integers less than (7)10¹⁷, one per line. The end of the input is indicated by a line containing only 0.

For each positive integer in the input, output a line containing only an A to Z numeral representing the integer.
Do not choose a solution method whose time is exponential in the number of digits!