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() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int i, sum = 0, a[36] = {0};
  string s;
  cin >> s;
  for (i = 0; i < s.size(); i++) {
    if (s[i] >= 'a') {
      a[s[i] - 'a' + 10]++;
    } else {
      a[s[i] - '0']++;
    }
  }
  for (i = 0; i < 36; i++) {
    sum += (a[i] * a[i]);
  }
  cout << sum;
  return 0;
}