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 (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int factorize(long long n, long long p[], int m[])
{
	int k;
	long long i;
	for (i = 2, k = 0; i * i <= n; i++) {
		if (n % i != 0) continue;
		p[k] = i;
		for (m[k] = 0; n % p[k] == 0; m[k]++, n /= p[k]);
		k++;
	}
	if (n > 1) {
		p[k] = n;
		m[k++] = 1;
	}
	return k;
}

long long count_divisors(long long n)
{
	int i, k, m[100];
	long long p[100], ans = 1;
	k = factorize(n, p, m);
	for (i = 0; i < k; i++) ans *= m[i] + 1;
	return ans;
}

int main()
{
	long long N;
	scanf("%lld", &N);
	
	long long K, n, ans = 0;
	for (K = 2; K * K < N; K++) {
		if (N % K != 0) continue;
		for (n = N; n % K == 0; n /= K);
		if (n % K == 1) ans++;
	}
	printf("%d\n", ans + count_divisors(N - 1));
	fflush(stdout);
	return 0;
}