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>
#define ll long long
using namespace std;

ll n, ans;

bool check(int x){
    ll t = n;
    while(t%x == 0) t /= x;
    return t%x == 1;
}

int main()
{
    cin >> n;
    ll m = n-1;
    for(int i = 2; (ll)i*i <= m; i++){
        if(m%i) continue;
        ans++;
        if((ll)i*i != m) ans++;
    }
    if(m > 1) ans++;
    for(int i = 2; (ll)i*i <= n; i++){
        if(n%i) continue;
        ans += check(i);
        if((ll)i*i != n) ans += check(n/i);
    }
    cout << ans+1 << endl;

    return 0;
}