Hyper operator a⁽ⁿ⁾b
is a family of operators, where each level n

operator is defined as iteration of the previous level (n - 1)
operator. It is defined recursively as follows:

<!-- MATH
$a^{(n)}b = \left { \begin{array}{cc} b+1, & \mbox{if } n=0 \a, & \mbox{if } n=1, b=0 \0, & \mbox{if } n=2,b=0 \1, & \mbox{if } n \ge 3, b = 0 \a^{{(n-1)}(a{(n)}(b-1))} & \mbox{if } n \ge 1, b \ge 1, a \ge 0 \end{array} \right.$
-->
a⁽ⁿ⁾b = WIDTH="20" HEIGHT="126" ALIGN="MIDDLE" BORDER="0"
SRC="/ntujudge/problemdata/635-1.png"
ALT="$ \left\{\vphantom{ \begin{array}{cc} b+1, & \mbox{if } n=0 \\ a, & \mbox{if } ...
...n-1)}(a^{(n)}(b-1)) & \mbox{if } n \ge 1, b \ge 1, a \ge 0 \end{array} }\right.$"> WIDTH="330" HEIGHT="126" ALIGN="MIDDLE" BORDER="0"
SRC="/ntujudge/problemdata/635-2.png"
ALT="$ \begin{array}{cc} b+1, & \mbox{if } n=0 \\ a, & \mbox{if } n=1, b=0 \\ 0, & ...
... \\ a^{(n-1)}(a^{(n)}(b-1)) & \mbox{if } n \ge 1, b \ge 1, a \ge 0 \end{array}$">

With some algebraic manipulations, it can be shown that the operators for n =
1, 2 and 3 correspond respectively to addition, multiplication and exponentiation; as given below:

<!-- MATH
$a^{(1)}b = a + {1+1+1+ \cdots + 1} = a + b$
-->
a⁽¹⁾b = a + 1+1+1+ ^... +1 = a + b

<!-- MATH
$a^{(2)}b = a + a + a + \cdots + a = a \times b$
-->
a⁽²⁾b = a + a + a + ^... + a = a×b

<!-- MATH
$a^{(3)}b = a \times a \times a \times \cdots \times a = a^{b}$
-->
a⁽³⁾b = a×a×a× ^... ×a = a^b

For values of n > 3
, a⁽ⁿ⁾b

typically corresponds to a (very) tall exponentiation tower even for small values of a
and b
, which makes it difficult or even impossible to calculate exactly. Nevertheless, as they are being iterations (or iterations of iterations of ...
) of exponents, their modulus can still be calculated by means of modular hyper-exponentiation (a form of iterated modular exponentiation). Your task is to compute <!-- MATH
$a^{(n)}b \bmod m$
-->
a⁽ⁿ⁾b mod m

, given the values of a
, b
, m
and n
.