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;
long long n;
int main()
{
    scanf("%lld",&n);
    if(n==2) {printf("1\n");return 0;}
    long long m;
    int cnt=0;
    for(long long i=2;i*i<=n;i++)
    {
        m=n;
        if(m%i) continue;
        while(m%i==0) m=m/i;
        if(m%i==1) cnt++;
    }
    n--;
    for(long long j=2;j*j<=n;j++)
    {
        if(n%j==0)
        {
            if(j*j==n) cnt++;
            else cnt+=2;
        }
    }
    printf("%d\n",cnt+2);
    return 0;
}
