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 <iostream>
#include <cstdint>
#include <unordered_map>
#include <algorithm>
#include <vector>

using i64 = std::int64_t;
using uint = std::uint32_t;

std::unordered_map<i64, int> primeFactor(i64 n) {
    std::unordered_map<i64, int> ret;
    for (i64 i = 2; i * i <= n; i++) {
        while (n % i == 0) {
            ret[i]++;
            n /= i;
        }
    }
    if (n != 1)
        ret[n] = 1;

    return ret;
}

i64 exp(i64 x, i64 n) {
    i64 ret = 1;
    while (n != 0) {
        if (n % 2 == 1)
            ret *= x;

        x *= x;
        n /= 2;
    }

    return ret;
}

int main() {
	i64 n;
	std::cin >> n;

    i64 ans = 1;
    for (auto&& p : primeFactor(n - 1))
        ans *= p.second + 1;

    auto map = primeFactor(n);
    std::vector<std::pair<i64, int>> pfacts(map.begin(), map.end());

    for (uint i = 1; i < (1 << map.size()); ++i) {
        int k = 1;
        int pow = 100000;
        for (int j = 0; j < map.size(); ++j) {
            if ((i >> j) & 1 == 1) {
                k *= pfacts[j].first;
                pow = std::min(pow, pfacts[j].second);
            }
        }

        i64 a = exp(k, pow);
        if (a == n)
            continue;

        for (int j = 1; j <= pow; ++j) {
            if (pow % j != 0)
                continue;

            if ((n / a - 1) % exp(k, j) == 0)
                ++ans;
        }
    }

    std::cout << ans << std::endl;
}