#include <bits/stdc++.h>
using namespace std;
int powmod(long long b, int p, int m) {
  long long r = 1;
  for (int i = (1 << 30); i; i >>= 1) {
    r *= r;
    r %= m;
    if (p & i) {
      r *= b;
      r %= m;
    }
  }
  return (int)r;
}
int main() {
  int p, total = 0;
  scanf("%d", &p);
  for (int x = (1); x < (p); x++) {
    int t = 1, i = 1;
    while (i <= p - 2) {
      t = t * x % p;
      if (t % p == 1) break;
      i++;
    }
    if (i > p - 2 && (t * x) % p == 1) {
      total++;
    }
  }
  printf("%d", total);
  return 0;
}

### Prompt

Your challenge is to write 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;
int powmod(long long b, int p, int m) {
  long long r = 1;
  for (int i = (1 << 30); i; i >>= 1) {
    r *= r;
    r %= m;
    if (p & i) {
      r *= b;
      r %= m;
    }
  }
  return (int)r;
}
int main() {
  int p, total = 0;
  scanf("%d", &p);
  for (int x = (1); x < (p); x++) {
    int t = 1, i = 1;
    while (i <= p - 2) {
      t = t * x % p;
      if (t % p == 1) break;
      i++;
    }
    if (i > p - 2 && (t * x) % p == 1) {
      total++;
    }
  }
  printf("%d", total);
  return 0;
}

```