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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    String A = sc.next();
    StringBuilder sb = new StringBuilder();
        sb.append("#");
        for (char c: A.toCharArray())sb.append(c+"#"); //ダミー文字を突っ込む
        char[] s = sb.toString().toCharArray();
        int N = s.length;
        long max = 0;
        
        long[] rad = new long[N];
        for (int i=0, j=0; i<N;) {
            while(0 <= i-j && i+j < N && s[i-j]==s[i+j]) j++; //rad[i]を素直に求める
            rad[i] = j;
            max = Math.max(max, j); // 最大値は記憶しておく
            // 計算しなくても求められる部分を走査
            int k;
            for (k=1; 0 <= i-k && i+k < N && rad[i-k] < rad[i]-k; k++) {
                rad[i+k] = rad[i-k];
            }
            i+=k; // 次のinteresting position
            j = Math.max(j-k, 0); // interesting position は rad[i]-kと同じか長い回文を持つので考慮させる
        }
    long kaibun = 0;
    for(int i = 0; i < N; i++) {
      kaibun += rad[i];
    }
    long n = A.length();
    long ans = 1 + ((n - 1) * n) / 2 - kaibun;
    System.out.println(ans);
  }
}