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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
//#include <stdlib.h>
//#include <string.h>
//#include <stdbool.h>
//#include <limits.h>
//#include <math.h>

int main() {
  long n;
  scanf("%ld", &n);
  if (n == 2) {
    printf("1\n");
    return 0;
  }
  long ans = 2;
  for (long i=2; i*i<=n; i++) {
    if (i*i == n-1)
      ans++;
    else if ((n-1)%i == 0)
      ans += 2;
    if (n%i != 0)
      continue;
    long p = n / i, m = n;
    while (m%i == 0)
      m /= i;
    if (m%i == 1)
      ans++;
    m = n;
    while (m%p == 0)
      m /= p;
    if (m%p == 1 && i!=p)
      ans++;
  }
  printf("%ld\n", ans);
  return 0;
}