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>

typedef long long ll;
using namespace std;

int main() {
    ll n;
    cin >> n;
    if (n == 2) {
        cout << 1 << endl;
        return 0;
    }
    ll ans = 2;
    vector<ll> nums;
    for (ll i = 2; i * i <= n; i++) {
        if (n % i == 0) {
            nums.emplace_back(i);
            nums.emplace_back(n / i);
        }
        if ((n - 1) % i == 0) {
            ans += 2;
        }
    }
    for (auto num: nums) {
        ll t = n;
        while (t % num == 0) {
            t /= num;
        }
        if (t % num == 1) {
            ans++;
        }
    }
    cout << ans << endl;
    return 0;
}