#include <bits/stdc++.h>
using namespace std;
const int BUFFER_SIZE = 1 * 1024;
const double EPS = 10e-6;
const int LIMIT = 100;
const int INF = 1000000000;
char buffer[BUFFER_SIZE];
char* read_line() {
  int ch, size = 0;
  while ((ch = getchar()) != '\n') buffer[size++] = ch;
  buffer[size] = '\0';
  return buffer;
}
long long modpow(long long b, int p, long long m) {
  long long mod = 1;
  while (p) {
    if (p % 2 == 1) mod = (mod * b) % m;
    b = (b * b) % m;
    p /= 2;
  }
  return mod;
}
int gcd(int a, int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
int main(int argc, char* argv[]) {
  int count = 0;
  long long p;
  cin >> p;
  int phi = p - 1;
  for (int i = 1; i < phi; i++)
    if (gcd(phi, i) == 1) count++;
  if (p == 2) count++;
  cout << count << endl;
  return EXIT_SUCCESS;
}

### Prompt

Your task is to create 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 int BUFFER_SIZE = 1 * 1024;
const double EPS = 10e-6;
const int LIMIT = 100;
const int INF = 1000000000;
char buffer[BUFFER_SIZE];
char* read_line() {
  int ch, size = 0;
  while ((ch = getchar()) != '\n') buffer[size++] = ch;
  buffer[size] = '\0';
  return buffer;
}
long long modpow(long long b, int p, long long m) {
  long long mod = 1;
  while (p) {
    if (p % 2 == 1) mod = (mod * b) % m;
    b = (b * b) % m;
    p /= 2;
  }
  return mod;
}
int gcd(int a, int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
int main(int argc, char* argv[]) {
  int count = 0;
  long long p;
  cin >> p;
  int phi = p - 1;
  for (int i = 1; i < phi; i++)
    if (gcd(phi, i) == 1) count++;
  if (p == 2) count++;
  cout << count << endl;
  return EXIT_SUCCESS;
}

```