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;

int main() {
    int64_t n;
    cin >> n;
    vector<int64_t> d;
    for (int64_t i = 1; i * i <= n; i++) {
	if (n % i == 0) {
	    d.push_back(i);
	    if (i * i < n) d.push_back(n / i);
	}
    }
    int64_t ans = 0;
    for (int64_t k : d) {
	if (k == 1) continue;
	int64_t cache = n;
	while (cache % k == 0) cache /= k;
	if (cache % k == 1) ans++;
    }
    for (int64_t i = 1; i * i <= n - 1; i++) {
	if ((n - 1) % i == 0) {
	    if (i > 1) ans++;
	    if (i * i < n - 1) ans++;
	}
    }
    cout << ans << endl;
}
