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 (Clang 10.0.0) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<stdbool.h>
#define rep(i,l,r) for(int i=l;i<r;i++)
#define MOD 1000000007

typedef long long ll;

int main() {
    ll n;
    scanf("%lld", &n);
    ll ans = 2;
    rep(k, 2, sqrt(n) + 1) {
        ll m1 = n, m2 = n;
        if (k * k == n - 1) ans++;
        else if (n - 1 % k == 0) ans += 2;
        if (k * k == n)ans++;
        else if (n % k == 0) {
            while (m1 % k == 0) m1 /= k;
            while (m2 % n / k == 0) m2 /= n / k;
            if (m1 % k == 1)ans++;
            if (m2 % k == 1)ans++;
        }
    }
    
    printf("%lld", ans);

    return 0;
}