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<iostream>
#include<cmath>
#include<algorithm>
#include<iomanip>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<map>
#include<set>
#include<fstream>
#include<sstream>
#include<queue>
#include<iterator>
#include<stack>

using namespace std;

#define st first
#define nd second
#define FOR(i,a,b) for(int i=a;i<b;i++)
#define REV(a) reverse(a.begin(),a.end())
#define SORT(a) sort(a.begin(),a.end())
#define inf (int)1e9
#define linf (long long)1e18

typedef long long ll;
typedef short int si;
typedef pair<int,int> pii;
typedef pair<ll,int> pli;
typedef vector<int> vi;
typedef vector<pii> vpii;
typedef vector<pli> vpli;

string s;
priority_queue<string> ans;
int main() {
	cin>>s;
	ans.push("");
	for(int i=0;i<s.size();i++){
		string a;
		a=s.substr(0,i+1);
		char c=s[i];
    priority_queue<string> use;
    while(!use.empty()) use.pop();
		for(int j=0;j<ans.size();j++){

			string b;
			b=ans.top();
			b=b+c;
			ans.pop();
			use.push(b);
		}
		ans=use;
		for(int j=0;j<i;j++){
			string b,c;
			b=c=a.substr(j,i-j+1);
			REV(c);
			if(b==c)continue;
			ans.push(a.substr(0,j)+c);
		}

	}
	ans.push(s);
	ll q=1;
	s=ans.top();
	ans.pop();
	while(!ans.empty()){
		 if(ans.top()!=s){
		 q++;
		 s=ans.top();
		}
		ans.pop();
	}
	cout<<q<<endl;

  return 0;
}