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>
#pragma optimize("O2")
using namespace std;
const long long MAXN = 2000 + 5;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  string s;
  cin >> s;
  long long n = s.size();
  map<char, long long> c;
  for (long long i = 0; i < n; ++i) {
    c[s[i]]++;
  }
  long long ans = 0;
  for (auto u : c) ans += u.second * u.second;
  cout << ans;
  return 0;
}