Problem:
Score : 600 points

Problem Statement

Given is a positive integer N.

We will choose an integer K between 2 and N (inclusive), then we will repeat the operation below until N becomes less than K.

Operation: if K divides N, replace N with N/K; otherwise, replace N with N-K.

In how many choices of K will N become 1 in the end?

Constraints

2 \leq N \leq 10^{12}

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of choices of K in which N becomes 1 in the end.

Sample Input 1

6

Sample Output 1

3

There are three choices of K in which N becomes 1 in the end: 2, 5, and 6.

In each of these choices, N will change as follows:

When K=2: 6 \to 3 \to 1

When K=5: 6 \to 1

When K=6: 6 \to 1

Sample Input 2

3141

Sample Output 2

13

Sample Input 3

314159265358

Sample Output 3

9

Solution in Pascal (FPC 2.6.2) (judge verdict: Accepted)

function check(N,x:int64):Boolean;
begin
if x=1 then check:=False else begin
while N mod x=0 do N:=N div x;
check:=N mod x=1;
end;
end;
function calc(N,M:int64):Longint;
var cnt:Longint;i:int64;
begin
i:=1;
while i*i<=M do begin
if M mod i=0 then begin
if check(N,i)then inc(cnt);
if(i*i<M)and(check(N,M div i))then inc(cnt);
end;
inc(i);
end;
calc:=cnt;
end;
var N:int64;
begin
read(N);
writeln(calc(N,N)+calc(N,N-1));
end.