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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  string s;
  cin >> s;
  map<char, long long> count;
  for (long long i = 0; i < (long long)s.size(); ++i) {
    map<char, long long>::iterator find = count.find(s[i]);
    if (find != count.end())
      ++find->second;
    else
      count.insert({s[i], 1});
  }
  long long res = 0;
  for (map<char, long long>::iterator it = count.begin(); it != count.end();
       ++it) {
    res += it->second * it->second;
  }
  cout << res;
  return 0;
}