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>
#include<iostream>
#include<vector>
#include<cmath>
using namespace std;
typedef long long ll;

vector<ll> getDivisor(ll n)
{
    vector<ll> ret;

    for (ll i = 2; i <= sqrt(n) + 1; i++) {
        if (n % i == 0) {
            ret.push_back(i);
            if (n != i * i) ret.push_back(n / i);
        }
    }
    ret.push_back(n);

    return ret;
}
int main(int, char**)
{
    ll n;

    cin >> n;

    vector<ll> divisors1 = getDivisor(n);
    vector<ll> divisors2 = getDivisor(n-1);

    ll ans = divisors2.size();
    ll tmp;
    for (auto x : divisors1) {
        tmp = n;
        while (tmp % x == 0) tmp /= x;
        if (tmp % x == 1) ans++;
    }

    cout << ans << endl;

    return 0;

}