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 <string>
#include <algorithm>
#include <vector>
#include <iomanip>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdlib>
#include <map>
#include <iomanip>
#include <set>
using namespace std;
#define pie 3.141592653589793238462643383279
#define int long long
#define mod 100000007
#define INF 1145141919364364
#define all(vec) vec.begin(),vec.end()
#define P pair<int,int>
#define S second
#define F first
int gcd(int x, int y) {
	if (y == 0)return x;
	return gcd(y, x%y);
}
int lcm(int x, int y) {
	return x / gcd(x, y)*y;
}
bool prime(int x) {
	for (int i = 2; i <= sqrt(x); i++) {
		if (x%i == 0)return false;
	}
	return true;
}
int kai(int x) {
	if (x == 0)return 1;
	return kai(x - 1) * x % mod;
}
int mod_pow(int x, int y, int mood) {
	int res = 1;
	while (y > 0) {
		if (y & 1) {
			res = res * x % mood;
		}
		x = x * x % mood;
		y >>= 1;
	}
	return res;
}
int comb(int x, int y) {
	if (y > x)return 0;
	return kai(x)* mod_pow(kai(x - y), mod - 2, mod) % mod * mod_pow(kai(y), mod - 2, mod) % mod;
}

string s;
int cnt[26];
int ans;

signed main() {
	cin >> s;
	ans = comb((int)s.size(), 2);
	for (int i = 0; i < (int)s.size(); i++)cnt[s[i] - 'a']++;
	for (int i = 0; i < 26; i++) {
		if (cnt[i])ans -= comb(cnt[i], 2);
	}
	cout << ans + 1 << endl;
}