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>
int main() {
  char str[100005];
  scanf("%s", &str);
  int leng = strlen(str);
  long long int ctr = 0, alfa[30] = {0}, num[12] = {0};
  for (int i = 0; i < leng; i++) {
    if (str[i] >= 'a' && str[i] <= 'z') {
      alfa[str[i] - 'a']++;
    } else {
      num[str[i] - '0']++;
    }
  }
  for (int i = 0; i < 26; i++) {
    ctr += pow(alfa[i], 2);
  }
  for (int i = 0; i < 10; i++) {
    ctr += pow(num[i], 2);
  }
  printf("%lld\n", ctr);
  return 0;
}