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 <bits/stdc++.h>
using namespace std;
#define LL long long

LL n, cnt = 0;

bool check (int x) {
	LL m = n;
	while (m % x == 0) m /= x;
	return m % x == 1;
}

int main () {
	cin >> n;
	LL sq = sqrt (n);
	for (LL i = 2; i <= sq; i ++) {
		if (n % i == 0) {
			if (check (i)) cnt ++;
			if (i != n / i && check (n / i)) cnt ++;
		}
	}
	n --;
	if (n > sq) {
		cnt ++;
		for (LL i = 2; i <= sq; i ++) {
			if (n % i == 0) {
				cnt ++;
				if (i != n / i) cnt ++;
			}
		}
	}
	cout << cnt + 1;
	return 0;
}