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;
char s[200000];
long long al[27];
long long nu[15];
int main() {
  scanf("%s", s);
  int len = strlen(s);
  memset(nu, 0, sizeof(nu));
  memset(al, 0, sizeof(al));
  for (int i = 0; i < len; i++) {
    if (s[i] >= 'a' && s[i] <= 'z') {
      al[s[i] - 'a']++;
    } else {
      nu[s[i] - '0']++;
    }
  }
  long long sum = 0;
  for (int i = 0; i < 26; i++) {
    sum += al[i] * al[i];
  }
  for (int i = 0; i < 10; i++) {
    sum += nu[i] * nu[i];
  }
  printf("%I64d\n", sum);
  return 0;
}