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 ch[100005];
  long long p[40];
  memset(p, 0, sizeof(p));
  scanf("%s", ch);
  int len = strlen(ch);
  for (int i = 0; i < len; ++i) {
    if (ch[i] >= '0' && ch[i] <= '9') {
      p[ch[i] - '0']++;
    } else {
      p[ch[i] - 'a' + 10]++;
    }
  }
  long long sum = 0;
  for (int i = 0; i < 40; ++i) sum += p[i] * p[i];
  printf("%lld\n", sum);
  return 0;
}