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() {
  string s;
  cin >> s;
  long long ans = 0;
  for (char i = 'a'; i <= 'z'; ++i) {
    long long cnt = 0;
    for (int j = 0; j < s.length(); ++j) {
      if (i == s[j]) ++cnt;
    }
    ans += (cnt * cnt);
  }
  for (char i = '0'; i <= '9'; ++i) {
    long long cnt = 0;
    for (int j = 0; j < s.length(); ++j) {
      if (i == s[j]) ++cnt;
    }
    ans += (cnt * cnt);
  }
  printf("%lld", ans);
  return 0;
}