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;
struct node {
  int data;
  long int sum;
  struct node *left, *right;
  struct node* next;
};
struct node* newNode(int data) {
  struct node* Node = new (struct node);
  Node->data = data;
  Node->left = Node->right = Node->next = NULL;
  return (Node);
}
long long count_arr[36];
void solveTestCase() {
  string str;
  cin >> str;
  int n = str.size();
  for (int i = 0; i < 36; i++) {
    count_arr[i] = 0;
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    if (str[i] >= '0' && str[i] <= '9') {
      count_arr[str[i] - '0']++;
    } else {
      count_arr[str[i] - 'a' + 10]++;
    }
  }
  for (int i = 0; i < 36; i++) {
    ans += (count_arr[i] * count_arr[i]);
  }
  printf("%I64d\n", ans);
  ;
}
int main() {
  solveTestCase();
  return 0;
}