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;
long long sum = 0, temp;
char s[100001];
int finder(int i) {
  int kol = 1;
  for (int j = 0; s[j] != 0; ++j) {
    if (j != i && s[j] == s[i]) {
      kol++;
      s[j] = 3;
    }
  }
  return kol;
}
int main() {
  gets(s);
  for (int i = 0; s[i] != 0; ++i) {
    if (s[i] != 3) {
      temp = finder(i);
      sum = sum + temp * temp;
      s[i] = 3;
    }
  }
  cout << sum;
  return 0;
}