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

int main() {
    ll N;
    cin >> N;
    ll ans = 0;
    ll r = sqrt(N);
    for (ll i = 2; i <= r; i++) {
        ll t = N;
        while (t%i == 0) t /= i;
        if (t%i == 1) ans++;
    }
    
    vector<ll> div;
    for (ll i = 1; i <= r; i++) {
        if ((N-1)%i == 0) {
            div.push_back(i);
            if (i*i != N-1) div.push_back((N-1)/i);
        }
    }
    for (ll k : div) {
        if (k > r) ans++;
    }
    ans++;
    cout << ans << endl;
    
    return 0;
}