#include <bits/stdc++.h>
using namespace std;
const int Max = 100008;
int a[Max];
char ch[502][502];
int dp_row[502] = {}, dp_col[502];
long long bigmod(long long b, long long p, long long m) {
  if (p == 0) return 1;
  if (p % 2 == 0) {
    long long temp = bigmod(b, p / 2, m);
    return (temp * temp) % m;
  }
  return (bigmod(b, p - 1, m) * (b % m)) % m;
}
int main() {
  int p, sum = 0;
  cin >> p;
  for (long long i = 1; i < p; i++) {
    bool ok = true;
    for (long long j = 1; j <= p - 2; j++)
      if ((bigmod(i, j, p)) - 1 == 0) {
        ok = false;
        break;
      }
    if (ok && (bigmod(i, p - 1, p) - 1 == 0)) sum++;
  }
  cout << sum;
  return 0;
}

### 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 Max = 100008;
int a[Max];
char ch[502][502];
int dp_row[502] = {}, dp_col[502];
long long bigmod(long long b, long long p, long long m) {
  if (p == 0) return 1;
  if (p % 2 == 0) {
    long long temp = bigmod(b, p / 2, m);
    return (temp * temp) % m;
  }
  return (bigmod(b, p - 1, m) * (b % m)) % m;
}
int main() {
  int p, sum = 0;
  cin >> p;
  for (long long i = 1; i < p; i++) {
    bool ok = true;
    for (long long j = 1; j <= p - 2; j++)
      if ((bigmod(i, j, p)) - 1 == 0) {
        ok = false;
        break;
      }
    if (ok && (bigmod(i, p - 1, p) - 1 == 0)) sum++;
  }
  cout << sum;
  return 0;
}

```