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<bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)


int main() {
	long long n;
	cin >> n;
	long long r = 0;
	long long t2 = n - 1;
	for(long long k = 2; k <=1000000; k++) {
		long long t = n;
		while(t % k == 0) t /= k;
		if (t % k == 1)  {
			r++;
		}
	}

	for(long long d = 1; d <= 1000000; d++) {
		if (t2 % d == 0) {
			long long k = t2 / d;
			if (k > 1000000) r++;
		}
	}
	if (n > 1000000) {
		r++;
	}
	cout << r << endl;


}
