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;
long long const mod = 1000000007;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
ifstream in("txt.in");
ofstream out("txt.out");
int main() {
  cout << fixed << setprecision(12);
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  string s;
  cin >> s;
  int n = s.size();
  sort(s.begin(), s.end());
  long long sum = 1;
  long long cnt = s.size();
  for (int i = 1; i < n; ++i) {
    if (s[i] == s[i - 1])
      ++sum;
    else {
      cnt += sum * (sum - 1);
      sum = 1;
    }
  }
  cout << cnt + sum * (sum - 1) << '\n';
  return 0;
}