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: Accepted)

#include <iostream>
#include <fstream>
#include <vector>
#include <set>
#include <map>
#include <cstring>
#include <string>
#include <cmath>
#include <cassert>
#include <ctime>
#include <algorithm>
#include <sstream>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <cstdlib>
#include <cstdio>
#include <iterator>
#include <functional>
#include <bitset>

#ifdef LOCAL
#define eprintf(...) fprintf(stderr,__VA_ARGS__)
#else
#define eprintf(...)
#endif

#define TIMESTAMP(x) eprintf("["#x"] Time : %.3lf s.\n", clock()*1.0/CLOCKS_PER_SEC)
#define TIMESTAMPf(x,...) eprintf("[" x "] Time : %.3lf s.\n", __VA_ARGS__, clock()*1.0/CLOCKS_PER_SEC)

using namespace std;

#ifdef LOCAL
static struct __timestamper {
  string what;
  __timestamper(const char* what) : what(what){};
  __timestamper(const string& what) : what(what){};
  ~__timestamper(){
    TIMESTAMPf("%s", what.data());
	}
} __TIMESTAMPER("end");
#else 
struct __timestamper {};
#endif

typedef long long ll;
typedef long double ld;

char s[210000];

int main() {
#ifdef LOCAL
  freopen("b.in", "r", stdin);
  freopen("b.out", "w", stdout);
#endif

  while (scanf("%s", s) == 1) {
    vector<int> cnt(26);
    ll ans = 1;
    for (int i = 0; s[i]; i++) {
      int id = s[i] - 'a';
      ans += i - cnt[id];
      cnt[id]++;
    }
    printf("%lld\n", ans);
  }

  return 0;
}