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;
const double pi = acos(-1.0);
void display_vect(vector<int> vect) {
  for (int i = 0; i < vect.size(); i++) printf("%d ", vect[i]);
  printf("\n");
}
void display_arr(int arr[], int n) {
  for (int i = 0; i < n; i++) printf("%d ", arr[i]);
  printf("\n");
}
void boost() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
void input_vect(vector<int>& vect) {
  for (int i = 0; i < vect.size(); i++) scanf("%d", &vect[i]);
}
void input_arr(int arr[], int n) {
  for (int i = 0; i < n; i++) scanf("%d", &arr[i]);
}
void solve() {
  string s;
  cin >> s;
  unordered_map<char, long long int> mp;
  for (int i = 0; i < s.size(); i++) {
    mp[s[i]]++;
  }
  long long int sum{};
  for (auto itr = mp.begin(); itr != mp.end(); itr++) {
    sum += pow(itr->second, 2);
  }
  cout << sum << endl;
}
int main() {
  solve();
  return 0;
}