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;
typedef long long ll;

ll n, ans;

int main(){
    cin >> n;
    for (ll i=2; i * i <= n; i++){
        if (n % i) continue;
        ll n2 = n;
        while (n2 % i == 0) n2 /= i;
        if (n2 % i == 1) ans++;
    }
    ll n2 = n - 1;
    for (ll i=1; i * i <= n2; i++){
        if (n2 % i) continue;
        ans += 2;
        if (i * i == n2) ans--;
    }
    // ans += (n > 2) + 1;
    cout << ans;
    return 0;
}