#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 10, M = N, INF = 0x3f3f3f3f, mod = 1e9 + 7,
                base = 131;
long long n, t, ans;
long long a[N], s[N];
long long phi(long long n) {
  long long res, a;
  res = a = n;
  for (long long i = 2; i <= a / i; i++) {
    if (a % i) continue;
    res = res / i * (i - 1);
    while (a % i == 0) a /= i;
  }
  if (a > 1) res = res / a * (a - 1);
  return res;
}
signed main() {
  long long p;
  cin >> p;
  cout << phi(p - 1);
  return 0;
}

### Prompt

Please formulate a cpp solution to 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;
const long long N = 2e5 + 10, M = N, INF = 0x3f3f3f3f, mod = 1e9 + 7,
                base = 131;
long long n, t, ans;
long long a[N], s[N];
long long phi(long long n) {
  long long res, a;
  res = a = n;
  for (long long i = 2; i <= a / i; i++) {
    if (a % i) continue;
    res = res / i * (i - 1);
    while (a % i == 0) a /= i;
  }
  if (a > 1) res = res / a * (a - 1);
  return res;
}
signed main() {
  long long p;
  cin >> p;
  cout << phi(p - 1);
  return 0;
}

```