#include <bits/stdc++.h>
using namespace std;
int gcd(int a, int b) { return (b == 0 ? a : gcd(b, a % b)); }
int lcm(int a, int b) { return ((a * b) / gcd(a, b)); }
long long pw(long long b, long long p) {
  if (!p) return 1;
  long long sq = pw(b, p / 2);
  sq *= sq;
  if (p % 2) sq *= b;
  return sq;
}
int sd(long long x) { return x < 10 ? x : x % 10 + sd(x / 10); }
int n, m;
string str1, str2;
char C;
int main() {
  int p, aa = 0;
  cin >> p;
  if (p == 2) {
    cout << 1 << endl;
    return 0;
  }
  p--;
  for (int i = 1; i < p; i++)
    if (gcd(p, i) == 1) aa++;
  cout << aa << endl;
}

### 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 gcd(int a, int b) { return (b == 0 ? a : gcd(b, a % b)); }
int lcm(int a, int b) { return ((a * b) / gcd(a, b)); }
long long pw(long long b, long long p) {
  if (!p) return 1;
  long long sq = pw(b, p / 2);
  sq *= sq;
  if (p % 2) sq *= b;
  return sq;
}
int sd(long long x) { return x < 10 ? x : x % 10 + sd(x / 10); }
int n, m;
string str1, str2;
char C;
int main() {
  int p, aa = 0;
  cin >> p;
  if (p == 2) {
    cout << 1 << endl;
    return 0;
  }
  p--;
  for (int i = 1; i < p; i++)
    if (gcd(p, i) == 1) aa++;
  cout << aa << endl;
}

```