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 C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main(){
ll N;
cin>>N;
ll ca=0;
ll cb=0;
ll cc=0;
ll N1=N;
if(N==2){
cout<<1<<endl;}
else{
for(long long i=1;i*i<=N-1;i++){
if((N-1)%i==0){
ca+=2;}
if(i*i==N-1){
ca--;}}
ca--;
for(long long i=2;i*i<=N;i++){
if(N%i==0){
while(N!=1){
if(N%i==0){
cb++;
N/=i;}
else{
cc++;
break;}}
if(cc==1){
cb=0;}
else{
break;}}
N=N1;}
cc=0;
for(long long i=1;i*i<=cb;i++){
if(cb%i==0){
cc+=2;}
if(i*i==cb){
cc--;}}
if(cc!=0){
cc--;}
for(long long i=2;i*i<=N;i++){
if(N%i==0){
while(N%i==0){
N/=i;}
if((N-1)%i==0){
ca++;}}
N=N1;}
ca++;
cout<<ca+cc<<endl;}}