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: Runtime Error)

#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i = 0; i < (n); ++i)
#define all(x) (x).begin(), (x).end()
#define dbg(x) cerr << #x << ": " << x << endl

bool dfs(ll i, ll k) {
  bool res = false;
  if(k == 1) return true;
  if(dfs(i, k - i)) res = true;
  if(k % i == 0)
    if(dfs(i, k / i)) res = true;
  return res;
}

int main() {
  ll n;
  cin >> n;
  int cnt = 0;
  for(ll i = 2; i * i < n; i++) {
    if(dfs(i, n)) cnt++;
  }
  cnt += 2;
  cout << cnt << endl;
  return 0;
}