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() {
  char s[1000000];
  long long a[37];
  long long n;
  int i = 0;
  while (gets(s) != NULL) {
    n = 0;
    for (i = 0; i < 37; ++i) a[i] = 0;
    i = 0;
    while (s[i] != '\0') {
      if (s[i] >= '0' && s[i] <= '9')
        a[s[i] - '0']++;
      else if (s[i] >= 'a' && s[i] <= 'z')
        a[s[i] - 'a' + 10]++;
      i++;
    }
    for (i = 0; i < 37; ++i) n += (a[i] * a[i]);
    cout << n << endl;
  }
  return 0;
}