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 Rust (1.15.1) (judge verdict: Accepted)

use std::io;
use std::cmp;


// read a line(without line feed)
fn read_line() -> String {
    let mut in_str = String::new();
    io::stdin().read_line(&mut in_str).unwrap();
    return in_str.trim().to_string();
}

// read wods on a line
pub fn read_words_in_line() -> Vec<String> {
    let s = read_line();
    return s.split_whitespace().map(|x| x.to_string()).collect::<Vec<String>>()
}

// read type t on a line
pub fn read_t_in_line<T : std::str::FromStr>() -> Vec<T> {
    let mut v : Vec<T> = Vec::new();
    for e in read_words_in_line().iter().map(|x| x.parse::<T>()) {
        if let Ok(r) = e {
            v.push(r);
        }
        else{
            return Vec::new();
        }
    }
    return v;
}

// sort slice
pub fn sort<T : std::cmp::Ord>(slice : &mut [T]) {
    slice.sort();
}
// sort slice by reverse order
pub fn sort_rev<T : std::cmp::Ord>(slice : &mut [T]) {
    slice.sort_by(|x, y| y.cmp(x));
}

pub fn min<T : std::cmp::Ord>(a : T, b : T) -> T{
    return cmp::min(a, b);
}

pub fn max<T : std::cmp::Ord>(a : T, b : T) -> T{
    return cmp::max(a, b);
}

fn main() {
    let s     = read_line().into_bytes();
    let n = s.len() as i64;
    let mut freq = vec![0 as i64 ; 26];
    for b in s {
        freq[(b - ('a' as u8)) as usize] += 1;
    }

    let mut tot = n * (n - 1)/ 2 + 1;
    for f in freq {
        tot -= f * (f - 1) / 2;
    }

    println!("{}", tot);

}