The Ackermann function is a classic example of a recursive function, notable especially because it is not a primitive recursive function. It grows very quickly in value, as does the size of its call tree.



The Ackermann function is usually defined as follows:







A
(
m
,
n
)
=


{



n
+
1



if 

m
=
0




A
(
m
−
1
,
1
)



if 

m
>
0

 and 

n
=
0




A
(
m
−
1
,
A
(
m
,
n
−
1
)
)



if 

m
>
0

 and 

n
>
0.








{\displaystyle A(m,n)={\begin{cases}n+1&{\mbox{if }}m=0\\A(m-1,1)&{\mbox{if }}m>0{\mbox{ and }}n=0\\A(m-1,A(m,n-1))&{\mbox{if }}m>0{\mbox{ and }}n>0.\end{cases}}}






Its arguments are never negative and it always terminates.



Task

Write a function which returns the value of 



A
(
m
,
n
)


{\displaystyle A(m,n)}

. Arbitrary precision is preferred (since the function grows so quickly), but not required.



See also

 Conway chained arrow notation for the Ackermann function.


ackermann(m, n) := if integerp(m) and integerp(n) then ackermann[m, n] else 'ackermann(m, n)$
 
ackermann[m, n] := if m = 0 then n + 1
                   elseif m = 1 then 2 + (n + 3) - 3
                   elseif m = 2 then 2 * (n + 3) - 3
                   elseif m = 3 then 2^(n + 3) - 3
                   elseif n = 0 then ackermann[m - 1, 1]
                   else ackermann[m - 1, ackermann[m, n - 1]]$
 
tetration(a, n) := if integerp(n) then block([b: a], for i from 2 thru n do b: a^b, b) else 'tetration(a, n)$
 
/* this should evaluate to zero */
ackermann(4, n) - (tetration(2, n + 3) - 3);
subst(n = 2, %);
ev(%, nouns);