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)

#pragma region template 1.0
#pragma region def
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<ll> vi;
typedef pair<ll, ll> ii;
#define REP(i, n) for (ll i = 0; i < (n); ++i)
#define REP1(i, n) for (ll i = 1; i <= (n); ++i)
#define OUT(x) cout << (x) << endl;
#define OUTA(a) \
    REP(i, (a).size()) { cout << (a[i]) << (i == (a).size() - 1 ? "\n" : " "); }
#define ALL(a) (a).begin(), (a).end()
#define SORT(a) sort(ALL(a))
#define RSORT(a) \
    SORT(a);     \
    reverse(ALL(a))

const ll INF = 1e18;
const ll MOD = 1e9 + 7;
#pragma endregion enddef

int main()
{
    ll N;
    cin >> N;
    ll ans = 0;
    for (ll K = 1; K * K <= N; ++K)
    {
        if (N % K == 0)
        {
            ll n = N;
            while (K > 1 and n % K == 0)
            {
                n /= K;
            }
            if (n % K == 1)
            {
                ++ans;
            }
            n = N;
            ll k = N / K;
            while (n % k == 0)
            {
                n /= k;
            }
            if (k > K and n % k == 1)
            {
                ++ans;
            }
        }
        if ((N - 1) % K == 0)
        {
            if (K > 1)
            {
                ++ans;
            }
            if ((N - 1) / K > K)
            {
                ++ans;
            }
        }
    }
    OUT(ans);
}
#pragma endregion template