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 cadbcdddda
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1.000000000000000
 0.000000000000000
 0.450000000000000
 0.960000000000000
 1.000000000000000
 0.800000000000000
 1.000000000000000
 0.080000000000000
 0.840000000000000
 0.000000000000000
 0.800000000000000
 1.000000000000000
 0.000000000000000
 0.786096256684492
 1.000000000000000
 0.438502673796791

#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
template <typename T, typename TT>
inline ostream &operator<<(ostream &os, const pair<T, TT> &t) {
  return os << t.first << " " << t.second;
}
template <typename T>
inline ostream &operator<<(ostream &os, const vector<T> &t) {
  for (auto i : t) os << i << " ";
  return os;
}
template <typename T>
inline istream &operator>>(istream &is, vector<T> &v) {
  for (T &t : v) is >> t;
  return is;
}
template <typename T1, typename T2>
inline istream &operator>>(istream &is, pair<T1, T2> &t) {
  is >> t.first >> t.second;
  return is;
}
const long long mod = 1e9 + 7;
void solve() {
  string s;
  cin >> s;
  long long n = s.size();
  vector<vector<vector<long long>>> v(
      26, vector<vector<long long>>(n, vector<long long>(26)));
  vector<long long> cv(26);
  for (int i = 0; i < n; i++) {
    cv[s[i] - 'a']++;
    for (int j = i + 1; j < n; j++) {
      v[s[i] - 'a'][j - i][s[j] - 'a']++;
      v[s[j] - 'a'][i - j + n][s[i] - 'a']++;
    }
  }
  long double ans = 0;
  for (int i = 0; i < 26; i++) {
    long double c = cv[i], mx = 0;
    for (int j = 1; j < n; j++) {
      long double x = 0, sum = 0;
      auto &cnt = v[i][j];
      for (auto &y : cnt) {
        if (y == 1) x++;
        sum += y;
      }
      if (!sum) continue;
      mx = max(mx, x / sum);
    }
    ans += c * mx;
  }
  printf("%.10Lf\n", ans / n);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  ;
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}
