Problem: There is a given string S consisting of N symbols. Your task is to find the number of ordered pairs of integers i and j such that

1. 1 ≤ i, j ≤ N

2. S[i] = S[j], that is the i-th symbol of string S is equal to the j-th.

Input

The single input line contains S, consisting of lowercase Latin letters and digits. It is guaranteed that string S in not empty and its length does not exceed 105.

Output

Print a single number which represents the number of pairs i and j with the needed property. Pairs (x, y) and (y, x) should be considered different, i.e. the ordered pairs count.

Examples

Input

great10


Output

7


Input

aaaaaaaaaa


Output

100

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
string to_string(char s) { return string(1, s); }
string to_string(string s) { return "\"" + s + "\""; }
template <typename T>
string to_string(vector<T>);
template <typename U, typename V>
string to_string(pair<U, V> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename T>
string to_string(vector<T> v) {
  string res = "[";
  for (int i = 0; i < (int)v.size() - 1; i++) res += to_string(v[i]) + ", ";
  if (!v.empty()) res += to_string(v.back());
  res += "]";
  return res;
}
char s[100100];
long long int f[26];
long long int f1[10];
int main() {
  long long int tot = 0;
  ;
  scanf("%s", s);
  for (int i = 0; s[i] != '\0'; i++) {
    if (s[i] >= 48 && s[i] <= 57)
      f1[s[i] - '0']++;
    else
      f[s[i] - 'a']++;
  }
  for (int i = 0; i < 26; i++) {
    tot += (f[i] * f[i]);
  }
  for (int i = 0; i < 10; i++) {
    tot += (f1[i] * f1[i]);
  }
  printf("%lld\n", tot);
}