#include <bits/stdc++.h>
using namespace std;
// Input
long long G;
long long N, A[1 << 17];
string S;
// DP
long long Cost[1 << 17][17];
long long sum1L[17][1 << 17];
long long sum1R[17][1 << 17];
long long sum2L[17][17][1 << 17];
long long sum2R[17][17][1 << 17];
long long dp[1 << 17];
long long GetLeft(int mask, int grp, int pos) {
if (pos == 0) return 0;
long long sum = 0;
for (int i = 0; i < G; i++) {
if (((mask >> i) & 1) == 0) continue;
sum += 2LL * sum1L[i][pos - 1];
}
sum += 1LL * sum1L[grp][pos - 1];
return sum;
}
long long GetRigt(int mask, int grp, int pos) {
if (pos == 0) return 0;
long long sum = 0;
for (int i = 0; i < G; i++) {
if (((mask >> i) & 1) == 0) continue;
sum += 2LL * sum1R[i][pos + 1];
}
sum += 1LL * sum1R[grp][pos + 1];
return sum;
}
long long GetLeftSum(int mask, int grp, int pos) {
long long ret = 0;
for (int i = 0; i < G; i++) {
if (((mask >> i) & 1) == 0) continue;
ret += sum2L[grp][i][pos];
}
ret += sum2L[grp][grp][pos];
return ret;
}
long long GetRigtSum(int mask, int grp, int pos) {
long long ret = 0;
for (int i = 0; i < G; i++) {
if (((mask >> i) & 1) == 0) continue;
ret += sum2R[grp][i][pos];
}
ret += sum2R[grp][grp][pos];
return ret;
}
long long GetCost(int mask, int grp) {
// Step A. Binary Search
int cl = 0, cr = N, cm;
for (int i = 0; i < 20; i++) {
cm = (cl + cr) / 2;
long long v1 = GetLeft(mask, grp, cm);
long long v2 = GetRigt(mask, grp, cm);
if (v1 < v2) { cl = cm; }
else { cr = cm; }
}
// Step B. Brute Force
long long Return = (1LL << 60);
for (int i = cm - 1; i <= cm + 1; i++) {
if (i < 0 || i >= N - 1) continue;
long long val1 = GetLeftSum(mask, grp, i);
long long val2 = GetRigtSum(mask, grp, i + 1);
Return = min(Return, val1 + val2);
}
return Return;
}
int main() {
// Step 1. Input
cin >> S;
N = S.size();
for (int i = 0; i < (int)S.size(); i++) {
A[i] = (S[i] - 'A');
G = max(G, A[i] + 1);
}
if (N <= 2) {
cout << 0 << endl;
return 0;
}
// Step 2. Precount 1
for (int i = 0; i < N; i++) {
sum1L[A[i]][i] += 1;
sum1R[A[i]][i] += 1;
}
for (int i = 0; i < G; i++) {
for (int j = 1; j <= N - 1; j++) sum1L[i][j] += sum1L[i][j - 1];
for (int j = N - 2; j >= 0; j--) sum1R[i][j] += sum1R[i][j + 1];
}
// Step 3. Precount 2
for (int i = 0; i < G; i++) {
for (int j = 0; j < G; j++) {
long long cur1 = 0;
long long sum1 = 0;
for (int k = 0; k < N; k++) {
if (A[k] == i) sum1 += cur1;
sum2L[i][j][k] = sum1;
if (A[k] == j && i == j) cur1 += 1;
if (A[k] == j && i != j) cur1 += 2;
}
long long cur2 = 0;
long long sum2 = 0;
for (int k = N - 1; k >= 0; k--) {
if (A[k] == i) sum2 += cur2;
sum2R[i][j][k] = sum2;
if (A[k] == j && i == j) cur2 += 1;
if (A[k] == j && i != j) cur2 += 2;
}
}
}
// Step 4. Precount 3
for (int i = 0; i < (1 << G); i++) {
for (int j = 0; j < G; j++) {
if ((i >> j) & 1) continue;
Cost[i][j] = GetCost(i, j);
}
}
/*for (int i = 0; i < (1 << G); i++) {
for (int j = 0; j < G; j++) printf("% 4lld", Cost[i][j]);
printf("\n");
}*/
// Step 5. DP
for (int i = 0; i < (1 << G); i++) dp[i] = (1LL << 60);
dp[0] = 0;
for (int i = 0; i < (1 << G); i++) {
for (int j = 0; j < G; j++) {
if ((i >> j) & 1) continue;
dp[i + (1 << j)] = min(dp[i + (1 << j)], dp[i] + Cost[i][j]);
}
}
// Step 6. Output
printf("%.12lf\n", 0.5 * dp[(1 << G) - 1]);
return 0;
}
