Problem:
Score : 500 points

Problem Statement

You have a string A = A_1 A_2 ... A_n consisting of lowercase English letters.

You can choose any two indices i and j such that 1 \leq i \leq j \leq n and reverse substring A_i A_{i+1} ... A_j.

You can perform this operation at most once.

How many different strings can you obtain?

Constraints

1 \leq |A| \leq 200,000

A consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

A

Output

Print the number of different strings you can obtain by reversing any substring in A at most once.

Sample Input 1

aatt

Sample Output 1

5

You can obtain aatt (don't do anything), atat (reverse A[2..3]), atta (reverse A[2..4]), ttaa (reverse A[1..4]) and taat (reverse A[1..3]).

Sample Input 2

xxxxxxxxxx

Sample Output 2

1

Whatever substring you reverse, you'll always get xxxxxxxxxx.

Sample Input 3

abracadabra

Sample Output 3

44

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define REP(i, n) for (int i = 0; i < n; ++i)
#define REPR(i, n) for (int i = n - 1; i >= 0; --i)
#define FOR(i, a, b) for (int i = a; i < b; ++i)
#define FORR(i, a, b) for (int i = b - 1; i >= a; --i)
#define SORT(v) sort(v.begin(), v.end())
#define SORTR(v) sort(v.rbegin(), v.rend())
#define REV(v) reverse(v.begin(), v.end())
#define ITER(itr, v) for (auto itr = v.begin(); itr != v.end(); ++itr)
#define LB(v, x) (lower_bound(v.begin(), v.end(), x) - v.begin())
#define UB(v, x) (upper_bound(v.begin(), v.end(), x) - v.begin())
#define SZ(v) (int)v.size()
using namespace std;
using ll = long long;
using P = pair<int, int>;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    string A;
    cin >> A;
    vector<ll> num_char(26);
    for (auto &c : A) {
        num_char[c - 'a']++;
    }
    ll res = 1;
    REP(i, 25) {
        FOR(j, i + 1, 26) {
            res += num_char[i] * num_char[j];
        }
    }
    cout << res << endl;

    return 0;
}