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 conversion(string p) {
  int o;
  o = atoi(p.c_str());
  return o;
}
string toString(int h) {
  stringstream ss;
  ss << h;
  return ss.str();
}
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
long long lcm(long long a, long long b) { return (a * (b / gcd(a, b))); }
long long toi(string p) {
  long long x;
  istringstream in(p);
  in >> x;
  return x;
}
int main() {
  set<char> S;
  map<char, long long> M;
  long long sum = 0, x;
  string a;
  cin >> a;
  for (int i = 0; i < a.size(); i++) {
    S.insert(a[i]);
    M[a[i]] += 1LL;
  }
  for (set<char>::iterator it = S.begin(); it != S.end(); it++) {
    x = M[*it];
    sum += ((x) * (x));
  }
  cout << sum << endl;
}