#include <iostream>
#include <vector>
#include <string>
using namespace std;
/*
#include <atcoder/all>
using namespace atcoder;
using mint = modint1000000007;
*/
#define all(x) (x).begin(),(x).end()
#define rep(i, n) for (int i = 0; i < (n); i++)
#define endl "\n"

#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

template <typename T> ostream &operator<<(ostream &os, const vector<T> &vec) {os << "["; for (const auto &v : vec) {os << v << ","; } os << "]"; return os;}
template <typename T, typename U> ostream &operator<<(ostream &os, const pair<T, U> &p) {os << "(" << p.first << ", " << p.second << ")"; return os;}
ll mod_pow(ll a, ll n, ll mod) { ll ret = 1; ll p = a % mod; while (n) { if (n & 1) ret = ret * p % mod; p = p * p % mod; n >>= 1; } return ret; }

const int mod = 1e9 + 7;

void solve() {
    string S; cin >> S;
    vector<vector<int>> acc(S.size() + 1, vector<int>(26));
    for (int i = 0; i < S.size(); i++) {
        for (int j = 0; j < 26; j++) {
            if (j == S[i] - 'A') {
                acc[i + 1][j] = acc[i][j] + 1;
            } else {
                acc[i + 1][j] = acc[i][j];
            }
        }
    }

    ll ans = 0;
    for (ll i = 0; i < S.size(); i++) {
        ll j = S[i] - 'A';
        ll prevc = acc[i][j];
        ll nextc = (S.size() - (i + 1)) - (acc[S.size()][j] - acc[i + 1][j]);
        ans += prevc * nextc;
    }
    cout << ans << endl;
}

int main() {
    #ifdef LOCAL_ENV
    cin.exceptions(ios::failbit);
    #endif
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout.setf(ios::fixed);
    cout.precision(16);
    
    solve();
}