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 <cstdio>
#include <iostream>
using namespace std;
typedef long long ll;
ll n;
bool check(ll x)
{
    ll t=n;
    while(t%x==0) t/=x;
    return t%x==1;
}
int sol(ll n)
{
    int re=0;
    for(int i=2;i<=n/i;++i) if(n%i==0)
    {
        if(check(i)) ++re;
        if((ll)i*i==n) continue;
        if(check(n/i)) ++re;
    }
    return re;
}
int main()
{
    scanf("%lld",&n);
    if(n==2)
    {
        puts("1");
        return 0;
    }
    int an=2;
    an+=sol(n);
    an+=sol(n-1);
    printf("%d\n",an);
    return 0;
}