#include <bits/stdc++.h>
using namespace std;
bool debug = false;
int n, m, k;
int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
long long ln, lk, lm;
int p;
bool ck(int x) {
  long long tmp = 1;
  for (int j = 1; j < p; j++) {
    if (tmp * (x - 1) % p == 0) {
      if (j == p - 1)
        return 1;
      else
        return 0;
    }
    tmp = (x * tmp + 1) % p;
  }
  return 0;
}
int main() {
  scanf("%d", &p);
  for (int x = 1; x < p; x++) {
    if (ck(x)) m++;
  }
  cout << m << endl;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
bool debug = false;
int n, m, k;
int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
long long ln, lk, lm;
int p;
bool ck(int x) {
  long long tmp = 1;
  for (int j = 1; j < p; j++) {
    if (tmp * (x - 1) % p == 0) {
      if (j == p - 1)
        return 1;
      else
        return 0;
    }
    tmp = (x * tmp + 1) % p;
  }
  return 0;
}
int main() {
  scanf("%d", &p);
  for (int x = 1; x < p; x++) {
    if (ck(x)) m++;
  }
  cout << m << endl;
  return 0;
}

```