#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
template <typename T, typename T1>
T amax(T &a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T amin(T &a, T1 b) {
  if (b < a) a = b;
  return a;
}
void OJ() {}
const long long N = 2e5 + 5;
long long solve() {
  long long p, cnt = 0;
  cin >> p;
  for (long long i = 1; i < p + 1; ++i) {
    long long f = 1;
    long long x = i;
    for (long long j = 1; j < p - 1; j++) {
      if ((x % p - 1 % p) % p == 0) f = 0;
      x = (x % p * i % p) % p;
    }
    if ((x % p - 1 % p) % p) f = 0;
    if (f) cnt++;
  }
  return cout << cnt, 0;
  ;
  return 0;
}
signed main() {
  OJ();
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  solve(), cout << '\n';
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
template <typename T, typename T1>
T amax(T &a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T amin(T &a, T1 b) {
  if (b < a) a = b;
  return a;
}
void OJ() {}
const long long N = 2e5 + 5;
long long solve() {
  long long p, cnt = 0;
  cin >> p;
  for (long long i = 1; i < p + 1; ++i) {
    long long f = 1;
    long long x = i;
    for (long long j = 1; j < p - 1; j++) {
      if ((x % p - 1 % p) % p == 0) f = 0;
      x = (x % p * i % p) % p;
    }
    if ((x % p - 1 % p) % p) f = 0;
    if (f) cnt++;
  }
  return cout << cnt, 0;
  ;
  return 0;
}
signed main() {
  OJ();
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  solve(), cout << '\n';
  return 0;
}

```