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 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#define pi 3.141592653589793238
#define int long long
using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

// mt19937 gen_rand_int(chrono::steady_clock::now().time_since_epoch().count());
// mt19937_64 gen_rand_ll(chrono::steady_clock::now().time_since_epoch().count());
// mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

long long power(long long a, long long b) {
    long long res = 1;
    while (b > 0) {
        if (b & 1)
            res = res * a ;
        a = a * a ;
        b >>= 1;
    }
    return res;
}

int a[305][305];
int pos[305][305];

signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);  
cout.tie(0);   

#ifndef ONLINE_JUDGE
    if(fopen("INPUT.txt","r"))
    {
    freopen ("INPUT.txt" , "r" , stdin);
    freopen ("OUTPUT.txt" , "w" , stdout);
    }
#endif  

    string a;
    cin>>a;
    int n=a.length();
    int mm[26]={0};
    for(int i=0;i<n;i++)
    {
        mm[a[i]-'a']++;
    }
    int ans=0;
    int tot=n;
    for(int i=0;i<n;i++)
    {
        tot--;
        mm[a[i]-'a']--;
        int z=mm[a[i]-'a'];
        ans+=tot-z;
    }
    cout<<ans+1;

}