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.

Examples

Input

6


Output

3


Input

3141


Output

13


Input

314159265358


Output

9
#include <algorithm>
#include <string.h>
#include <iostream>
#include <stdio.h>

using namespace std;

typedef long long LL;

int main() {
	LL n, res = 2;
	cin >> n;
	if (n == 2) res = 1;
	for (LL k = 2, i = n; k * k <= n; k++, i = n) {
		while (i >= k) i % k ? i %= k : i /= k;
		if (i == 1) res++;
		if ((n - 1) % k == 0 && k * k != (n - 1)) res++;
	}
	cout << res << endl;
	return 0;
}
