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 <math.h>


int main(void)
{
	//変数の宣言
	char a[200010];
	int alphabet[30];
	int i,j,k;
	long int len;
	long int ans;
	
	//データの読み込み
	scanf("%s",&a);
	
	
//	printf("nは%dです\n", n);
//	printf("データの読み込み終了\n");
	//実際の処理
	
	len=strlen(a);
	for(i=0;i<len;i++){
		alphabet[a[i]-'a']++;
	}
	
	ans=len*(len-1)/2+1;
	for(i=0;i<26;i++){
//		if(alphabet[i]!=0) printf("%cは%d個\n",'a'+i,alphabet[i]);
		ans=ans-(long)alphabet[i]*(alphabet[i]-1)/2;
	}
	
	
//	printf("計算部分終了\n");
	
	//出力
	printf("%d",ans);
	
//	printf("結果の出力終了\n");
	
	return 0;
}