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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
char a[200005];
int ne[200005];
void get_next(int m)
{
    int i,j;
    i=0;
    j=-1;
    ne[i]=-1;
    while(i<m)
    {
        if(j==-1||a[i]==a[j])
            ne[++i]=++j;
        else j=ne[j];
    }
    return ;
}
int main()
{
    int i,j,m,n;
    ios::sync_with_stdio(false);
    cin>>a;
    m=strlen(a);
    get_next(m);
    ll sum=0;
    for(i=0;i<=m;i++)
    {
        sum+=ne[i];
        //cout<<ne[i]<<endl;
    }
    if(sum==-1) {
        cout<<m*(m-1)/2<<endl;
    }
    else cout<<m*(m-1)/2-sum<<endl;
    return 0;
}