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 <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <iomanip>
#include <vector>
#include <string.h>
#include <string>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <math.h>
#include <algorithm>
#include <numeric>

using namespace std;

// マクロ&定数 ============================================================
typedef unsigned int uint;
typedef long long ll;
//typedef pair<int, int> P;

typedef vector<int> vint;
typedef vector<ll> vll;
typedef vector<double> vdouble;
typedef vector<bool> vbool;
typedef vector<string> vstring;

typedef vector<pair<int, int>> vpint;
typedef vector<pair<ll, ll>> vpll;
typedef vector<pair<double, double>> vpdouble;

typedef vector<vector<int>> vvint;
typedef vector<vector<ll>> vvll;
typedef vector<vector<double>> vvdouble;
typedef vector<vector<bool>> vvbool;

const int c_YET = -1;
const int INF = 1e9 + 1;
const ll LLINF = 1e17 + 1;
const int DX[9] = { 0,0,1,-1, 1, 1, -1, -1, 0 }; // 4;４近傍
const int DY[9] = { 1,-1,0,0, 1, -1, 1, -1, 0 }; // 8:８近傍 9:(0,0)を含む
const ll MOD = 1e9 + 7; //10^9 + 7
const ll MAX = 1e9;
const double PI = 3.14159265358979323846264338327950288;
//========================================================================


vll yakusuu(ll n) {
	vll 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);
}



ll N;
bool check(ll n)
{
	ll k = N;
	while (k >= n)
	{
		if (k % n == 0)
		{
			k /= n;
		}
		else
		{
			k %= n;
		}
	}

	if (k == 1)
	{
		return true;
	}
	else
	{
		return false;
	}
}

int main() {

	////==================================
	//cin.tie(nullptr);
	//ios_base::sync_with_stdio(false);
	//cout << fixed << setprecision(30);
	////==================================

	cin >> N;


	vll yaku_N = yakusuu(N);
	vll yaku_N1 = yakusuu(N - 1);



	ll ans = yaku_N1.size() - 1;


	for (int i = 1; i < yaku_N.size(); i++)
	{
		if (yaku_N[i] == 6)
		{
			int df = 0;
		}

		if (check(yaku_N[i]))
		{
			ans++;
		}
	}

	cout << ans;

}