#include <bits/stdc++.h>
using namespace std;
int prime;
long long ans;
int flag = 1;
long long mypow(long long x, long long p, long long mod) {
  long long ret = 1;
  x %= mod;
  while (p > 0) {
    if (p & 1) ret = (ret * x) % mod;
    p = p >> 1;
    x = (x * x) % mod;
  }
  return ret;
}
int main() {
  cin >> prime;
  for (int i = 1; i < prime; i++) {
    for (int j = 1; j < prime - 1; j++)
      if (mypow((long long)i, (long long)j, (long long)prime) - 1 == 0) {
        flag = 0;
        break;
      }
    if (flag) ans++;
    flag = 1;
  }
  printf("%I64d\n", ans);
  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;
int prime;
long long ans;
int flag = 1;
long long mypow(long long x, long long p, long long mod) {
  long long ret = 1;
  x %= mod;
  while (p > 0) {
    if (p & 1) ret = (ret * x) % mod;
    p = p >> 1;
    x = (x * x) % mod;
  }
  return ret;
}
int main() {
  cin >> prime;
  for (int i = 1; i < prime; i++) {
    for (int j = 1; j < prime - 1; j++)
      if (mypow((long long)i, (long long)j, (long long)prime) - 1 == 0) {
        flag = 0;
        break;
      }
    if (flag) ans++;
    flag = 1;
  }
  printf("%I64d\n", ans);
  return 0;
}

```