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;
const long long INFLL = (1LL << 62);
const int INF = (1 << 30), MOD = (int)1e9 + 7;
long long f(int n) { return (long long)n * n; }
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  string s;
  cin >> s;
  map<char, int> cnt;
  for (char& c : s) {
    cnt[c]++;
  }
  long long res = 0;
  for (char c = '0'; c <= '9'; c++) {
    res += (long long)f(cnt[c]);
  }
  for (char c = 'a'; c <= 'z'; c++) {
    res += (long long)f(cnt[c]);
  }
  cout << res << '\n';
  return 0;
}