#include <bits/stdc++.h>
using namespace std;
int N;
void init() { cin >> N; }
inline bool succ(int x) {
  int x_ = x;
  for (int i = 0; i < N - 2; ++i) {
    int y = ((x - 1 == 0) ? N : x - 1);
    if (y % N == 0) return false;
    x = (x * x_) % N;
  }
  x = (x * x_) % N;
  return 1;
}
void sol() {
  int ans = 0;
  for (int i = 1; i < N; ++i)
    if (succ(i)) ++ans;
  cout << ans << endl;
}
int main() {
  init();
  sol();
}

### 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;
void init() { cin >> N; }
inline bool succ(int x) {
  int x_ = x;
  for (int i = 0; i < N - 2; ++i) {
    int y = ((x - 1 == 0) ? N : x - 1);
    if (y % N == 0) return false;
    x = (x * x_) % N;
  }
  x = (x * x_) % N;
  return 1;
}
void sol() {
  int ans = 0;
  for (int i = 1; i < N; ++i)
    if (succ(i)) ++ans;
  cout << ans << endl;
}
int main() {
  init();
  sol();
}

```