Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

6

Sample Output 1

3

There are three choices of K in which N becomes 1 in the end: 2, 5, and 6.

In each of these choices, N will change as follows:

When K=2: 6 \to 3 \to 1

When K=5: 6 \to 1

When K=6: 6 \to 1

Sample Input 2

3141

Sample Output 2

13

Sample Input 3

314159265358

Sample Output 3

9

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<queue>
#include<stack>
#include<numeric>
#include<math.h>
using namespace std;
#define INF (1 << 30) - 1
#define LINF 1LL << 60
const int MOD = 1000000007;
using ll = long long;
using P = pair<int, int>;

vector<ll> divisors(ll n){ 
	vector<ll> res;
	for(ll i = 1; i * i <= n; ++i){
		if(n % i)continue;
		res.push_back(i);
		if(i * i != n)res.push_back(n / i);
	}
	return res;
}

int main(){
  ll n;
  cin >> n;
  ll cnt = 0;

  for(ll k : divisors(n)){
  	if(k == 1)continue;
  	ll tmp = n;
  	while(tmp % k == 0)tmp /= k; 
  	if(tmp % k == 1)++cnt;
  }

  cnt += divisors(n - 1).size() - 1;

  cout << cnt << endl;
  return 0;
}