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 rep(i,n) for(int i=0; i<(int)(n); i++)

using namespace std;
typedef long long LL;
const int Max_Num=1000010;

int main(){
  LL N;
  cin >> N;
  LL ans=0;
  for(LL K=2; K*K<=N; K++){
    if(N%K!=0) continue;
    LL Nc=N;
    while(Nc%K==0) Nc/=K;
    if(Nc%K==1) ans++;
  }
  LL div[Max_Num]={0};
  N--;
  for(LL K=2; K*K<=N; K++) while(N%K==0) N/=K, div[K]++;
  LL num=1;
  for(LL K=2; K<Max_Num; K++) num*=div[K]+1;
  if(N!=1) num*=2;
  ans+=num;
  cout << ans << endl;

  return 0;
}
