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;
int main(){
  long long N;
  cin >> N;
  set<long long> F;
  for (long long i = N - 1; i <= N; i++){
    for (long long j = 1; j * j <= i; j++){
      if (i % j == 0){
        F.insert(j);
        F.insert(i / j);
      }
    }
  }
  int count = 0;
  for (auto K : F){
    if (K == 1) continue;
    long long N2 = N;
    while (N2 % K == 0) N2 /= K;
    if (N2 % K == 1) count++;
  }
  cout << count;
}