#include using namespace std; struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_; #define FOR(i, begin, end) for(int i=(begin);i<(end);i++) #define REP(i, n) FOR(i,0,n) #define IFOR(i, begin, end) for(int i=(end)-1;i>=(begin);i--) #define IREP(i, n) IFOR(i,0,n) #define Sort(v) sort(v.begin(), v.end()) #define Reverse(v) reverse(v.begin(), v.end()) #define all(v) v.begin(),v.end() #define SZ(v) ((int)v.size()) #define Lower_bound(v, x) distance(v.begin(), lower_bound(v.begin(), v.end(), x)) #define Upper_bound(v, x) distance(v.begin(), upper_bound(v.begin(), v.end(), x)) #define Max(a, b) a = max(a, b) #define Min(a, b) a = min(a, b) #define bit(n) (1LL<<(n)) #define bit_exist(x, n) ((x >> n) & 1) #define debug(x) cout << #x << "=" << x << endl; #define vdebug(v) cout << #v << "=" << endl; REP(i_debug, v.size()){ cout << v[i_debug] << ","; } cout << endl; #define mdebug(m) cout << #m << "=" << endl; REP(i_debug, m.size()){ REP(j_debug, m[i_debug].size()){ cout << m[i_debug][j_debug] << ","; } cout << endl;} #define pb push_back #define f first #define s second #define int long long #define INF 1000000000000000000 template istream &operator>>(istream &is, vector &v){ for (auto &x : v) is >> x; return is; } template ostream &operator<<(ostream &os, vector &v){ for(int i = 0; i < v.size(); i++) { cout << v[i]; if(i != v.size() - 1) cout << endl; }; return os; } template void Out(T x) { cout << x << endl; } template void Ans(bool f, T1 y, T2 n) { if(f) Out(y); else Out(n); } using vec = vector; using mat = vector; using Pii = pair; using PiP = pair; using PPi = pair; using bools = vector; using pairs = vector; //int dx[4] = {1,0,-1,0}; //int dy[4] = {0,1,0,-1}; //char d[4] = {'D','R','U','L'}; const int mod = 1000000007; //const int mod = 998244353; #define Add(x, y) x = (x + (y)) % mod #define Mult(x, y) x = (x * (y)) % mod signed main(){ string S; cin >> S; int cnt = 0; int N = SZ(S); vec last(26, -1); REP(i, N){ if(last[S[i] - 'a'] == -1){ cnt += (i + 1) * (N - i); }else{ cnt += (i - last[S[i] - 'a']) * (N - i); } last[S[i] - 'a'] = i; } int tot = N * (N + 1) / 2; Out((double)cnt / tot); return 0; }