#include <bits/stdc++.h>
int x;
int pow(int a, int b) {
  int ret, rem;
  if (b == 1) return a % x;
  if (b % 2)
    rem = a % x;
  else
    rem = 1;
  ret = (pow(a, b / 2) % x);
  return (((ret % x) * (ret % x)) % x * (rem % x)) % x;
}
bool isprim(int number) {
  int i, a;
  bool flag;
  flag = 1;
  for (i = 1; i <= x - 2; i++) {
    a = (pow(number, i) - 1) % x;
    if (a > 0) continue;
    if (a == 0)
      flag = 0;
    else {
      while (a < 0) a += x;
      if (a % x == 0) flag = 0;
    }
  }
  a = (pow(number, i) - 1) % x;
  while (a < 0) a += x;
  if (((a) % x) == 0 && flag) return 1;
  return 0;
}
int main() {
  int i, answer = 0;
  scanf("%d", &x);
  for (i = 1; i < x; i++)
    if (isprim(i)) answer++;
  printf("%d\n", answer);
}

### Prompt

In cpp, your task is to solve the following problem:

The cows have just learned what a primitive root is! Given a prime p, a primitive root <image> is an integer x (1 ≤ x < p) such that none of integers x - 1, x2 - 1, ..., xp - 2 - 1 are divisible by p, but xp - 1 - 1 is. 

Unfortunately, computing primitive roots can be time consuming, so the cows need your help. Given a prime p, help the cows find the number of primitive roots <image>.

Input

The input contains a single line containing an integer p (2 ≤ p < 2000). It is guaranteed that p is a prime.

Output

Output on a single line the number of primitive roots <image>.

Examples

Input

3


Output

1


Input

5


Output

2

Note

The only primitive root <image> is 2.

The primitive roots <image> are 2 and 3.

### Response

```cpp
#include <bits/stdc++.h>
int x;
int pow(int a, int b) {
  int ret, rem;
  if (b == 1) return a % x;
  if (b % 2)
    rem = a % x;
  else
    rem = 1;
  ret = (pow(a, b / 2) % x);
  return (((ret % x) * (ret % x)) % x * (rem % x)) % x;
}
bool isprim(int number) {
  int i, a;
  bool flag;
  flag = 1;
  for (i = 1; i <= x - 2; i++) {
    a = (pow(number, i) - 1) % x;
    if (a > 0) continue;
    if (a == 0)
      flag = 0;
    else {
      while (a < 0) a += x;
      if (a % x == 0) flag = 0;
    }
  }
  a = (pow(number, i) - 1) % x;
  while (a < 0) a += x;
  if (((a) % x) == 0 && flag) return 1;
  return 0;
}
int main() {
  int i, answer = 0;
  scanf("%d", &x);
  for (i = 1; i < x; i++)
    if (isprim(i)) answer++;
  printf("%d\n", answer);
}

```