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 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;
}
