Expected Allowance

Hideyuki receives some 1000-yen bills each month from his father Ujisato as pocket money. On the first day of every month, the number of bills is decided as follows. Ujisato prepares $n$ dice, each with $m$ sides, and declares a cutback $k$. Hideyuki rolls all of the dice. The number of bills he receives is the sum of the rolled spots decreased by the cutback. Fortunately for Hideyuki, Ujisato always gives him at least one bill, even when the sum of the spots does not exceed the cutback. Each die shows a value from $1$ to $m$ on its sides, and every side is equally likely.

Write a program that computes the expected number of bills Hideyuki receives.

For example, when $n = 2$, $m = 6$ and $k = 3$, the number of bills is $\max(1, S - 3)$, where $S$ is the sum of the two dice. The probabilities that the number of bills is $1, 2, 3, 4, 5, 6, 7, 8, 9$ are $\frac{1}{36}+\frac{2}{36}+\frac{3}{36}$, $\frac{4}{36}$, $\frac{5}{36}$, $\frac{6}{36}$, $\frac{5}{36}$, $\frac{4}{36}$, $\frac{3}{36}$, $\frac{2}{36}$ and $\frac{1}{36}$, respectively. Therefore the expected value is $(\frac{1}{36}+\frac{2}{36}+\frac{3}{36})\times 1 + \frac{4}{36}\times 2 + \frac{5}{36}\times 3 + \frac{6}{36}\times 4 + \frac{5}{36}\times 5 + \frac{4}{36}\times 6 + \frac{3}{36}\times 7 + \frac{2}{36}\times 8 + \frac{1}{36}\times 9 = \frac{37}{9}$ (approximately $4.111111$).

The input is a sequence of lines, each of which contains three integers $n$, $m$ and $k$ in this order. They satisfy the following conditions:

• $1 \le n$
• $2 \le m$
• $0 \le k < nm$
• $nm \times m^n < 10^8$

The end of the input is indicated by a line containing three zeros.

For each input line, print the expected number of bills as an exact reduced fraction. Because every side of every die is equally likely, the expectation is a rational number; print it in the form $p/q$ where $q \ge 1$ and $\gcd(p, q) = 1$ (when the expectation is an integer, print it as $p/1$). Print no other characters.