Problem: Vasya and Kolya play a game with a string, using the following rules. Initially, Kolya creates a string s, consisting of small English letters, and uniformly at random chooses an integer k from a segment [0, len(s) - 1]. He tells Vasya this string s, and then shifts it k letters to the left, i. e. creates a new string t = sk + 1sk + 2... sns1s2... sk. Vasya does not know the integer k nor the string t, but he wants to guess the integer k. To do this, he asks Kolya to tell him the first letter of the new string, and then, after he sees it, open one more letter on some position, which Vasya can choose.

Vasya understands, that he can't guarantee that he will win, but he wants to know the probability of winning, if he plays optimally. He wants you to compute this probability. 

Note that Vasya wants to know the value of k uniquely, it means, that if there are at least two cyclic shifts of s that fit the information Vasya knowns, Vasya loses. Of course, at any moment of the game Vasya wants to maximize the probability of his win.

Input

The only string contains the string s of length l (3 ≤ l ≤ 5000), consisting of small English letters only.

Output

Print the only number — the answer for the problem. You answer is considered correct, if its absolute or relative error does not exceed 10 - 6.

Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if <image>

Examples

Input

technocup


Output

1.000000000000000


Input

tictictactac


Output

0.333333333333333


Input

bbaabaabbb


Output

0.100000000000000

Note

In the first example Vasya can always open the second letter after opening the first letter, and the cyclic shift is always determined uniquely.

In the second example if the first opened letter of t is "t" or "c", then Vasya can't guess the shift by opening only one other letter. On the other hand, if the first letter is "i" or "a", then he can open the fourth letter and determine the shift uniquely.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long maxn = 2e6 + 6;
const long long MOD = 1e9 + 7;
bool comparator(long long i, long long j) { return (i < j); }
long long power(long long x, long long i) {
  long long ans = 1;
  while (i > 0) {
    if (i & 1) ans = (ans * x) % MOD;
    i >>= 1;
    x = (x * x) % MOD;
  }
  return ans;
}
long long power(long long x, long long i, long long mod) {
  long long ans = 1;
  while (i > 0) {
    if (i & 1) ans = (ans * x) % mod;
    i >>= 1;
    x = (x * x) % mod;
  }
  return ans;
}
long long modInverse(long long x, long long mod) {
  return power(x, mod - 2, mod);
}
bool isPalindrome(string s) {
  long long limit = s.length() / 2;
  for (long long i = 0; i < limit; i++) {
    if (s[i] != s[s.length() - i - 1]) return 0;
  }
  return true;
}
long long gcd(long long x, long long y) {
  long long t;
  while (y != 0) {
    t = x % y;
    x = y;
    y = t;
  }
  return x;
}
bool isVowel(char ch) {
  if (ch == 'a' || ch == 'i' || ch == 'e' || ch == 'u' || ch == 'o' ||
      ch == 'y') {
    return true;
  }
  return false;
}
bool isPrime(long long n) {
  long long root = sqrt(n);
  for (long long i = 2; i <= root; i++)
    if (n % i == 0) return 0;
  return 1;
}
long double getDis(long double asdfg, long double qwert, long double gfdsa,
                   long double trewq) {
  return sqrt((asdfg - gfdsa) * (asdfg - gfdsa) +
              (qwert - trewq) * (qwert - trewq));
}
long double getDisSquare(long double asdfg, long double qwert,
                         long double gfdsa, long double trewq) {
  return (asdfg - gfdsa) * (asdfg - gfdsa) + (qwert - trewq) * (qwert - trewq);
}
bool isSquareHelper(long long asdfg, long long qwert, long long gfdsa,
                    long long trewq, long long x3, long long y3, long long x4,
                    long long y4) {
  long double d1, d2, d3, d4, d5, d6;
  d1 = getDisSquare(asdfg, qwert, gfdsa, trewq);
  d2 = getDisSquare(gfdsa, trewq, x3, y3);
  d3 = getDisSquare(x3, y3, x4, y4);
  d4 = getDisSquare(x4, y4, asdfg, qwert);
  if (d1 == d2 && d1 == d3 && d1 == d4) {
    d5 = getDisSquare(asdfg, qwert, x3, y3);
    d6 = getDisSquare(gfdsa, trewq, x4, y4);
    if (d5 == d6) {
      return 1;
    }
  }
  return 0;
}
bool isSquare(long long asdfg, long long qwert, long long gfdsa,
              long long trewq, long long x3, long long y3, long long x4,
              long long y4) {
  if (isSquareHelper(asdfg, qwert, gfdsa, trewq, x3, y3, x4, y4) ||
      isSquareHelper(asdfg, qwert, gfdsa, trewq, x4, y4, x3, y3) ||
      isSquareHelper(asdfg, qwert, x3, y3, gfdsa, trewq, x4, y4))
    return 1;
  else
    return 0;
}
bool isEqualateralTriangle(long long asdfg, long long qwert, long long gfdsa,
                           long long trewq, long long x3, long long y3) {
  long long d1 = getDisSquare(asdfg, qwert, gfdsa, trewq);
  long long d2 = getDisSquare(gfdsa, trewq, x3, y3);
  long long d3 = getDisSquare(x3, y3, asdfg, qwert);
  if (d1 == d2 && d1 == d3) {
    return 1;
  } else
    return 0;
}
bool isRightAngleTriangleHelper(long long asdfg, long long qwert,
                                long long gfdsa, long long trewq, long long x3,
                                long long y3) {
  long long d1 = getDisSquare(asdfg, qwert, gfdsa, trewq);
  long long d2 = getDisSquare(gfdsa, trewq, x3, y3);
  long long d3 = getDisSquare(x3, y3, asdfg, qwert);
  if (d2 == (d1 + d3)) {
    return 1;
  }
  return 0;
}
bool isRightAngleTriangle(long long asdfg, long long qwert, long long gfdsa,
                          long long trewq, long long x3, long long y3) {
  if (isRightAngleTriangleHelper(asdfg, qwert, gfdsa, trewq, x3, y3) ||
      isRightAngleTriangleHelper(gfdsa, trewq, x3, y3, asdfg, qwert) ||
      isRightAngleTriangleHelper(x3, y3, asdfg, qwert, gfdsa, trewq))
    return 1;
  return 0;
}
bool areCollinear(long long asdfg, long long qwert, long long gfdsa,
                  long long trewq, long long x3, long long y3) {}
string s;
long long n, b[30], dp[30];
vector<long long> g[100];
void getInput() {
  cin >> s;
  n = s.length();
}
void solve() {
  for (long long i = 0; i < n; i++) {
    g[s[i] - 'a'].push_back(i);
  }
  for (long long i = 0; i < 26; i++) {
    if (g[i].size() == 0) continue;
    long long mx = 0, cnt;
    for (long long j = 1; j <= n - 1; j++) {
      for (long long i = 0; i < 26; i++) b[i] = 0;
      for (auto k : g[i]) {
        b[s[(k + j) % n] - 'a']++;
      }
      cnt = 0;
      for (long long k = 0; k < 26; k++) {
        if (b[k] == 1) cnt++;
      }
      mx = max(cnt, mx);
    }
    dp[i] = mx;
  }
  long double ans = 0;
  for (long long i = 0; i < 26; i++) {
    ans += dp[i];
  }
  cout << fixed << setprecision(10) << (long double)ans / (long double)n
       << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) getInput(), solve();
  return 0;
}