//Sylwia Sapkowska
#include <bits/stdc++.h>
#pragma GCC optimize("O3", "unroll-loops")
using namespace std;
void __print(int x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << "'" << x << "'";}
void __print(const char *x) {cerr << '"' << x << '"';}
void __print(const string &x) {cerr << '"' << x << '"';}
void __print(bool x) {cerr << (x ? "true" : "false");}
template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ", "; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? ", " : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifdef LOCAL
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif
#define int long long
typedef pair<int, int> T;
const int oo = 1e18, oo2 = 1e9+7, K = 30;
const int mod = 998244353;
typedef long double ld;
void solve(){
string s; cin >> s;
int n = (int)s.size();
vector<int>used(K);
for (auto c: s) used[c-'A']++;
vector<int>scale(K);
int G = 0;
for (int i = 0; i<K; i++) if (used[i]) scale[i] = G++;
debug(G); //grupy [0, G-1]
vector<vector<int>>what(G);
for (int i = 0; i<n; i++) {
what[scale[s[i]-'A']].emplace_back(i);
}
vector<ld>dp((1<<G), oo);
dp[0] = 0;
vector left(n, vector<int>(G)), right(n, vector<int>(G)); //dla i-tego indeksu grupa g doklada left[i] elementow z lewej
for (int i = 0; i<n; i++){
for (int g = 0; g < G; g++){
if (scale[s[i]-'A'] == g) continue;
int t = (int)(lower_bound(what[g].begin(), what[g].end(), i)-what[g].begin());
left[i][g] = t;
right[i][g] = (int)what[g].size()-t;
}
}
vector<int>prev(G, -1);
vector pref(n, vector<int>(G)), suf(n, vector<int>(G));
vector lewo(n, vector<int>(G)), prawo(n, vector<int>(G));
for (int i = 0; i<n; i++){
int c = scale[s[i]-'A'];
if (prev[c] != -1) pref[i] = pref[prev[c]];
//pref[i][g] -> suma po left[j][g] po wszystkich j<=i, ze j jest w tej samej grupie co i
for (int g = 0; g < G; g++){
pref[i][g] += left[i][g];
}
prev[c] = i;
lewo[i] = prev;
}
prev.assign(G, -1);
for (int i = n-1; i>=0; i--){
int c = scale[s[i]-'A'];
if (prev[c] != -1) suf[i] = suf[prev[c]];
for (int g = 0; g < G; g++) suf[i][g] += right[i][g];
prev[c] = i;
prawo[i] = prev;
}
//2^{G-1} * G * (GlogN)
for (int mask = 0; mask < (1<<G); mask++){
for (int i = 0; i<G; i++){
if (mask&(1<<i)) continue;
//dokladam grupe i --> to sie wydarzy 2^{G-1} razy
int k = (int)what[i].size();
auto check = [&](int p){
ld L = (ld)p/2.0;
ld R = (ld)(k-p-1)/2.0;
for (int rep = 0; rep < G; rep++){
if (mask&(1<<rep)){
L += left[what[i][p]][rep];
R += right[what[i][p]][rep];
}
}
return L <= R;
};
auto total = [&](int p) -> ld {
//jesli osoba p jest ostatnia osoba ktora idzie z lewej strony, to koszt to?
//L = (0 + 1 + .. + p)/2 R = (0 + 1 + ... + (k-(p+1)-1))/2
ld L = (ld)(p*(p+1))/4.0;
ld R = (ld)(k-p-2)*(k-p-1)/4.0;
for (int rep = 0; rep < G; rep++){
if (mask&(1<<rep)){
if (p >= 0) {
L += pref[what[i][p]][rep];
debug(pref[what[i][p]][rep]);
}
if (p+1 < k){
R += suf[what[i][p+1]][rep];
debug(suf[what[i][p+1]][rep]);
}
}
}
return L+R;
};
//pewien prefiks przyjdzie z lewej, pewien sufiks z prawej
int l = 0, r = k-1, curr = -1;
while (r >= l){
int m = (l+r)/2;
if (check(m)){
curr = m;
l = m+1;
} else {
r = m-1;
}
}
dp[mask^(1<<i)] = min(dp[mask^(1<<i)], dp[mask] + total(curr));
}
}
cout << fixed << setprecision(10) << dp.back() << "\n";
}
int32_t main(){
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t = 1;
//cin >> t;
while (t--) solve();
return 0;
}
