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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

//#pragma GCC optimize(2)
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=2e6+5;
const int mod=1e9+7;
ll n,s,a[maxn],ss,val,ans;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n;
    for(int i=2;i*i<=n;i++){
        if(n%i==0) {
            a[++s]=i;
            if(i!=n/i) a[++s]=n/i;
        }
        if((n-1)%i==0){
            ans++;
            if(i!=(n-1)%i) ans++;
        }
    }
    ans++;
    a[++s]=n;
    for(int i=1;i<=s;i++){
        val=n;
        while(val%a[i]==0){
            val=val/a[i];
        }
        if((val-1)%a[i]==0) ss++;
    }
    cout<<ans+ss<<endl;
    return 0;
}