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)

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); i++)
#define all(x) (x).begin(), (x).end()
using ll = long long;
using namespace std;

int main() {
    ll N;
    cin >> N;
    int ans = 0;

    for(ll i=2;i*i<=(N-1);++i){
        if((N-1)%i != 0) continue; 
        ++ans;
        if(i*i != (N-1)) ++ans;
    }
    ++ans;
    for(ll i=2;i*i<=N;++i){
        ll tmpN = N; 
        if(N%i != 0) continue;
        while(tmpN%i==0) tmpN/=i;
        if(tmpN%i == 1) ++ans;
    }
    ++ans;
    cout << ans << endl;
}