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>
const long long maxn = 100000000 + 10;
using namespace std;
int main() {
  char a[100000 + 10];
  cin >> a;
  long long int size = 0;
  long long sum = 0;
  while (a[size] != 0) size++;
  map<char, int> m;
  for (int i = 0; i < size; i++) {
    map<char, int>::iterator itr = m.find(a[i]);
    if (itr == m.end()) {
      m.insert(make_pair(a[i], 1));
    } else {
      m[a[i]]++;
    }
  }
  map<char, int>::iterator itr = m.begin();
  while (itr != m.end()) {
    long long int k;
    k = itr->second;
    sum += (k * k);
    itr++;
  }
  cout << sum;
}