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;
void print(const vector<int> &v, bool endl = true) {
  cout << "[ ";
  for (int i = 0; i < v.size(); ++i) {
    cout << v[i] << " ";
  }
  cout << "]";
  if (endl) {
    cout << endl;
  }
}
void print(const string &s) { cout << s << endl; }
void print(const map<char, vector<int> > &m) {
  cout << "{" << endl;
  for (auto it = m.begin(); it != m.end(); it++) {
    cout << "  " << it->first << ": ";
    print(it->second, false);
    cout << "," << endl;
  }
  cout << "}" << endl;
}
using namespace std;
long long int variation(int m, int n) {
  if (m < n) {
    return 0;
  }
  long long int ret = 1;
  for (int i = m - n + 1; i <= m; ++i) {
    ret *= i;
  }
  return ret;
}
int main() {
  string s;
  while (cin >> s) {
    map<char, vector<int> > m;
    long long int ret = s.size();
    for (int i = 0; i < s.size(); ++i) {
      if (m.find(s[i]) == m.end()) {
        m[s[i]] = vector<int>(1, i);
      } else {
        m[s[i]].push_back(i);
      }
    }
    for (auto it = m.begin(); it != m.end(); it++) {
      ret += variation(it->second.size(), 2);
    }
    cout << ret << endl;
  }
  return 0;
}