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;
string s;
long long a[28] = {0};
long long b[11] = {0};
int main() {
  cin >> s;
  for (int i = 0; i < s.length(); ++i) {
    if (s[i] >= 'a' && s[i] <= 'z') a[s[i] - 'a'] += 1;
  }
  for (int i = 0; i < s.length(); ++i) {
    if (s[i] >= '0' && s[i] <= '9') b[s[i] - '0'] += 1;
  }
  long long ans = 0;
  for (int i = 0; i < 26; i++) {
    if (a[i] == 0) continue;
    ans += a[i] * a[i] * 1LL;
  }
  for (int i = 0; i < 10; i++) {
    if (b[i] == 0) continue;
    ans += b[i] * b[i] * 1LL;
  }
  cout << ans << endl;
  return 0;
}