#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const long long INF = 1e9 + 9;
const long long N = 500500;
long long exp(long long x, long long y, long long md) {
  long long res = 1;
  x = x % md;
  while (y) {
    if (y & 1) {
      res = (res * x) % md;
    }
    x = (x * x) % md;
    y = y >> 1;
  }
  return res;
}
bool primroot(long long x, long long p) {
  for (long long i = 1; i <= p - 2; i++) {
    if (exp(x, i, p) == 1) return false;
  }
  if (exp(x, p - 1, p) == 1) return true;
  return false;
}
signed main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.precision(10);
  cout << fixed;
  long long p, ans = 0;
  cin >> p;
  for (long long i = 1; i <= p - 1; i++) {
    if (primroot(i, p)) {
      ans++;
    }
  }
  cout << ans << '\n';
  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;
const long long MOD = 1e9 + 7;
const long long INF = 1e9 + 9;
const long long N = 500500;
long long exp(long long x, long long y, long long md) {
  long long res = 1;
  x = x % md;
  while (y) {
    if (y & 1) {
      res = (res * x) % md;
    }
    x = (x * x) % md;
    y = y >> 1;
  }
  return res;
}
bool primroot(long long x, long long p) {
  for (long long i = 1; i <= p - 2; i++) {
    if (exp(x, i, p) == 1) return false;
  }
  if (exp(x, p - 1, p) == 1) return true;
  return false;
}
signed main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.precision(10);
  cout << fixed;
  long long p, ans = 0;
  cin >> p;
  for (long long i = 1; i <= p - 1; i++) {
    if (primroot(i, p)) {
      ans++;
    }
  }
  cout << ans << '\n';
  return 0;
}

```