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 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(void) {
	char A[200000];
	long long int ans = 0, k, l;
	
	scanf("%s", A);
	ans++;
	for(k = 2; k <= strlen(A); k++) 
		ans += strlen(A)-k+1;
	// 奇数で左右対称
	for(k = 1, l = 1; k < strlen(A) - 1; k++) {
		if(A[k-l] == A[k+l]) {
			ans--;
			for(l = 2; k-l >= 0 && k+l <= strlen(A)-1; l++) {	
				if(A[k-l] == A[k+l])
					ans--;
				else
					break;
			}
			l = 1;
		}
	}
	// 偶数で左右対称
	for(k = 0, l = 0; k < strlen(A) - 1; k++) {
		if(A[k] == A[k+1]) {
			ans--;
			for(l = 1; k-l >= 0 && k+1+l <= strlen(A)-1; l++) {
				if(A[k-l] == A[k+1+l])
					ans--;
				else
					break;
			}
			l = 0;
		}
	}

	printf("%lld", ans);

	return 0;
}