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>
#define ff first
#define ss second
#define ll long long
#define rep(i, n) for (int i = 0; i < n; i++)
using namespace std;


int main() {
	ios::sync_with_stdio(false);	cin.tie(NULL);	cout.tie(NULL);

	ll n, final = 0;	cin >> n;

	for (ll i = 2; i * i <= n; i++) {
		ll x = n;
		while (x % i == 0)	x /= i;
		if (x % i == 1)		final++;
	}

	ll x = n - 1;

	for (ll i = 1; i * i <= x; i++)
		if (x % i == 0)		final++;

	cout << final + 1 << endl;

	return 0;
}