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;

vector<int64_t> divs(int64_t N){
    vector<int64_t> res;
    for(int64_t i=1; i*i <= N; i++) if(N%i == 0){
        res.push_back(i);
        if(i < N/i) res.push_back(N/i);
    }
    return res;
}

int main(){
    int64_t N;
    cin >> N;

    int64_t ans = divs(N-1).size()-1;
    auto D = divs(N);
    for(int64_t k : D) if(k != 1){
        int64_t n = N;
        while(n%k == 0) n /= k;
        if(n%k == 1) ans++;
    }
    cout << ans << endl;
    return 0;
}
