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;
//cout << 'H' << endl;
/*--------------------------------------------*/

map<ll, ll> prime_factor(ll N) {
    map< ll, ll > ret;
    for(ll i = 2; i * i <= N; i++) {
        while(N % i == 0) {
            ret[i]++;
            N /= i;
        }
    }
    if(N != 1) ret[N] = 1;
    return ret;
}

int main()
{
    ll N;
    cin >> N;
    ll ans = 1;
    for(auto p: prime_factor(N-1)){
        ans *= p.second + 1;
    }
    cout << (ans + 1) << endl;
}