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: Accepted)

#include <algorithm>
#include <cmath>
#include <complex>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <vector>
#include <functional>
#include <cstring>
#include <regex>
#include <random>
#include <cassert>
#include <stack>
using namespace std;

#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define repr(i, s, n) for (int i = (s); i < (int)(n); i++)
#define revrep(i, n) for (int i = (n) - 1; i >= 0; i--)
#define revrepr(i, s, n) for (int i = (n) - 1; i >= s; i--)
#define debug(x) cerr << #x << ": " << x << "\n"
#define popcnt(x) __builtin_popcount(x)

using ll = long long;
using P = pair<int, int>;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

template<class T>
istream& operator >>(istream &is, vector<T> &v) {
    for (int i = 0; i < (int)v.size(); i++) cin >> v.at(i);
    return is;
}

template<class T, class U>
ostream& operator <<(ostream &os, pair<T, U> p) {
    cout << '(' << p.first << ", " << p.second << ')';
    return os;
}

template<class T>
void print(const vector<T> &v, const string &delimiter) { rep(i, v.size()) cout << (0 < i ? delimiter : "") << v.at(i); cout << endl; }

template<class T>
void print(const vector<vector<T>> &vv, const string &delimiter) { for (const auto &v: vv) print(v, delimiter); }

vector<ll> divisor(ll n) {
    vector<ll> res;
    for (ll i = 1; i * i <= n; i++)
    {
        if (n % i == 0) {
            res.push_back(i);
            if (i != n / i) res.push_back(n / i);
        }
    }
    return res;
}

bool is_ok(ll n, ll k) {
    while (n % k == 0) n /= k;
    return n % k == 1;
}

int main() {
    ll n;
    cin >> n;
    auto div = divisor(n);
    int cnt = 0;
    for (ll d: div) {
        if (d == 1) continue;
        if (is_ok(n, d)) cnt++;
    }
    cnt += (int)divisor(n - 1).size() - 1;
    cout << cnt << endl;
}