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 INF = 1e6 + 10;
const int MaxN = 1e5 + 1;
int dx[] = {1, 1, -1, -1, 2, 2, -2, -2};
int dy[] = {2, -2, 2, -2, 1, -1, 1, -1};
using namespace std;
long long a[26], b[10];
int main(void) {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  memset(a, 0, sizeof a);
  memset(b, 0, sizeof b);
  string s;
  cin >> s;
  for (int i = 0; i < s.length(); i++)
    if (s[i] >= '0' && s[i] <= '9') {
      b[s[i] - '0']++;
    } else {
      a[s[i] - 'a']++;
    }
  long long ans = 0;
  for (int i = 0; i < 26; i++) ans += (a[i] * a[i]);
  for (int i = 0; i < 10; i++) ans += (b[i] * b[i]);
  cout << ans << endl;
  return 0;
}