#include <bits/stdc++.h>
using namespace std;
int n, nr;
int expo(int x, int p) {
  int sol = 1;
  while (p > 0) {
    if (p % 2 == 1) sol = (sol * x) % n;
    x = (x * x) % n;
    p = p / 2;
  }
  return sol;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    bool ok = true;
    for (int j = 1; j <= n - 2; j++) {
      int val = expo(i, j) - 1;
      if (val == 0) ok = false;
    }
    int val = expo(i, n - 1) - 1;
    if (val != 0) ok = false;
    if (ok) nr++;
  }
  printf("%d", nr);
  return 0;
}

### 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>
using namespace std;
int n, nr;
int expo(int x, int p) {
  int sol = 1;
  while (p > 0) {
    if (p % 2 == 1) sol = (sol * x) % n;
    x = (x * x) % n;
    p = p / 2;
  }
  return sol;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    bool ok = true;
    for (int j = 1; j <= n - 2; j++) {
      int val = expo(i, j) - 1;
      if (val == 0) ok = false;
    }
    int val = expo(i, n - 1) - 1;
    if (val != 0) ok = false;
    if (ok) nr++;
  }
  printf("%d", nr);
  return 0;
}

```