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: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define all(v) v.begin(), v.end()
using namespace std;
typedef long long ll;

vector<ll> divisor(ll n) {
  vector<ll> ret;
  for(ll i = 1; i * i <= n; i++) {
    if(n % i == 0) {
      ret.push_back(i);
      if(i * i != n) ret.push_back(n / i);
    }
  }
  sort(begin(ret), end(ret));
  return (ret);
}

int main() {
	ll n;
	cin >> n;
	auto p = divisor(n);
	auto q = divisor(n-1);
	set<ll> s;
	rep(i, p.size()) s.insert(p[i]);
	rep(i, q.size()) s.insert(q[i]);
	ll cnt = 0;
	for(ll t: s) {
		if(t==1) continue;
		ll tmp = n;
		if(tmp % t == 1) {
			tmp %= t;
			if(tmp == 1) cnt++;
		} else while(tmp > 0) {
			if(tmp==1) {
				cnt++;
				break;
			}
			tmp /= t;
		}
	}
	cout << cnt << endl;
	return 0;
}