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>

using namespace std;
typedef long long ll;
const int INF = 1e9;
const int MOD = 1e9 + 7;
const ll LINF = 1e18;

vector<pair<ll, ll>> prime_factor(ll n) {
    vector<pair<ll, ll>> res;

    for (ll i = 2; i * i <= n; ++i) {
        if (n % i != 0) continue;

        ll num = 0;

        while (n % i == 0) {
            ++num;
            n /= i;
        }

        res.push_back(make_pair(i, num));
    }

    if (n != 1) res.push_back(make_pair(n, 1));

    return res;
}

int main() {

    ll n; cin >> n;

    vector<pair<ll, ll>> pf = prime_factor(n);

    set<ll> se;

    for (auto p : pf) {
        for (int i = 2; i <= sqrt(p.first); ++i) {
            if (p.first % i == 1) {
                se.insert(i);
            }
        }
    }

    cout << se.size() * 2 + 3 << endl;
}