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 maxarr(vector<long long> &arr);
long long minarr(vector<long long> &arr);
long long gcd(long long a, long long b);
long long choose(long long n, long long k);
bool isprime(long long n);
void setprime(bool isprime[], long long n);
int main() {
  string str;
  cin >> str;
  map<char, long long> hm;
  for (long long i = 0; i < str.length(); i++) hm[str[i]]++;
  long long count = 0;
  for (map<char, long long>::iterator itr = hm.begin(); itr != hm.end();
       itr++) {
    count += (itr->second) * (itr->second);
  }
  cout << count << endl;
}
long long maxarr(vector<long long> &arr) {
  long long maxll = LONG_LONG_MIN;
  for (long long i = 0; i < arr.size(); i++) maxll = max(arr[i], maxll);
  return maxll;
}
long long minarr(vector<long long> &arr) {
  long long minll = LONG_LONG_MAX;
  for (long long i = 0; i < arr.size(); i++) minll = min(arr[i], minll);
  return minll;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long choose(long long n, long long k) {
  if (k == 0) return 1;
  return (n * choose(n - 1, k - 1)) / k;
}
bool isprime(long long n) {
  if (n == 2) return true;
  if (n % 2 == 0) return false;
  for (long long i = 3; i * i <= n; i += 2) {
    if (n % i == 0) return false;
  }
  return true;
}
void setprime(bool isprime[], long long n) {
  for (long long i = 0; i < n; i++) isprime[i] = true;
  isprime[0] = false;
  isprime[1] = false;
  for (long long i = 2; i < n; i++) {
    for (long long j = 2; i * j < n; j++) isprime[i * j] = false;
  }
}