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 <iostream>
#include <vector>
#include <algorithm>
#include <map>
#include <stack>
#include <set>
#include <queue>
#include <iomanip>
#include <math.h>

using namespace std;

long long mod = 1000000007;

long long binpow(long long x, long long k)
{
    if (k == 0)
    {
        return 1LL;
    }
    long long temp = binpow(x, k / 2);
    if (k % 2 == 1)
    {
        return (((temp * temp) % mod) * x) % mod;
    }
    else
    {
        return (temp * temp) % mod;
    }
}

long long modular_inverse(long long x)
{
    return binpow(x, mod - 2);
}

long long binomial(long long n, long long k)
{
    long long rez = 1;
    for (long long i = n - k + 1; i <= n; i++)
    {
        rez *= i;
        rez %= mod;
    }
    long long temp = 1;
    for (long long i = 1; i <= k; i++)
    {
        temp *= i;
        temp %= mod;
    }
    rez *= modular_inverse(temp);
    rez %= mod;
    return rez;
}

int fi(int a, vector<int> &p)
{
    if (p[a] != p[p[a]])
    {
        p[a] = fi(p[a], p);
    }
    return p[a];
}

void uni(int a, int b, vector<int> &p)
{
    int x = fi(a, p), y = fi(b, p);
    if (x != y)
    {
        p[x] = y;
    }
}

class SegmentTree
{
public:
    class Node
    {
    public:
        vector<int> value;
        Node *l, *r;
        int left, right;

        Node(int x, int y)
        {
            value = vector<int>(26, 0);
            l = nullptr;
            r = nullptr;
            left = x;
            right = y;
        }

        void update(int p, char c)
        {
            if (left > p || p > right)
            {
                return;
            }
            if (left == p && p == right)
            {
                for (int i = 0; i < 26; i++)
                {
                    value[i] = 0;
                }
                value[c - 'a'] = 1;
                return;
            }
            if (l == nullptr)
            {
                int mid = (left + right) / 2;
                l = new Node(left, mid);
                r = new Node(mid + 1, right);
            }
            l->update(p, c);
            r->update(p, c);
            for (int i = 0; i < 26; i++)
            {
                value[i] = l->value[i] + r->value[i];
            }
        }

        int query(int x, int y)
        {
            if (left > y || x > right)
            {
                return 0;
            }
            if (x <= left  && right <= y)
            {
                int rez = 0;
                for (int i = 0; i < 26; i++)
                {
                    if (value[i] != 0)
                    {
                        rez |= (1 << i);
                    }
                }
                return rez;
            }
            if (l == nullptr)
            {
                return 0;
            }
            return l->query(x, y) | r->query(x, y);
        }
    };

    Node *root;
    SegmentTree(int n)
    {
        root = new Node(0, n - 1);
    }

    void update(int p, char c)
    {
        root->update(p, c);
    }

    int query(int x, int y)
    {
        return root->query(x, y);
    }
};

int main()
{
    long long n;
    cin >> n;
    if (n == 2)
    {
        cout << "1\n";
        return 0;
    }
    long long temp;
    set<long long> st;
    st.insert(n);
    st.insert(n - 1);
    for (long long i = 2; i * i <= n; i++)
    {
        temp = n;
        while (temp % i == 0)
        {
            temp /= i;
        }
        if (temp % i == 1)
        {
            st.insert(i);
        }
        if ((n - 1) % i == 0 && n % ((n - 1) / i) != 0)
        {
            st.insert((n - 1) / i);
        }
    }
    cout << st.size() << '\n';
    return 0;
}