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 <algorithm>
#include <bitset>
#include <cassert>
#include <chrono>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <limits>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <ratio>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
#include <climits>
#define ll long long
#define ld long double
#define mp make_pair
#define pb push_back
#define in insert
#define vll vector<ll>
#define endl "\n"
#define pll pair<ll,ll>
#define f first
#define s second
#define FOR(i,a,b) for(int i=(a),_b=(b); i<=_b; i++)
#define int ll
#define sz(x) (ll)x.size()
#define all(x) (x.begin(),x.end())
using namespace std;

 
const ll INF = 1e12;
const ll N =(3e5+5); // TODO : change value as per problem
const ll MOD = 1e9+7;
void solve(){	
	string s;
	cin >> s;
	int n = (int)s.size();
	map<char,int> m;
	for(int i =0;i<n;i++){
		m[s[i]]++;
	}
	int cur = (n*(n-1))/2;
	for(char c = 'a';c<='z';c++){
		cur -= (m[c]*(m[c]-1))/2;
	}
	cout << cur + 1 << endl;
}
signed main(){
 
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
     // freopen(".in","r",stdin);freopen(".out","w",stdout);
    
     ll tt=1;   
     // cin >> tt;
    while(tt--){    
        solve();
    }    
}