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 gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long binpow(long long a, int n) {
  if ((!a) && (!n)) return 0;
  long long res = 1;
  while (n) {
    if (n & 1) res *= a;
    a *= a;
    n = n >> 1;
  }
  return res;
}
void InSort(int *s, int n) {
  int min;
  for (int i = 0; i < n; ++i) {
    min = i;
    for (int j = i + 1; j < n; ++j)
      if (s[j] < s[min]) min = j;
    swap(s[i], s[min]);
  }
}
void read(long long *s, int n) {
  for (int i = 0; i < n; ++i) scanf("%lld", &s[i]);
}
#pragma comment(linker, "/stack:64000000")
int let[32];
int num[10];
int main() {
  memset(let, 0, sizeof let);
  memset(num, 0, sizeof num);
  string str;
  cin >> str;
  for (int i = 0; i < str.length(); ++i) {
    if (str[i] <= '9' && str[i] >= '0')
      ++num[str[i] - '0'];
    else
      ++let[str[i] - 'a'];
  }
  long long ans = 0;
  for (int i = 0; i < 32; ++i) ans += let[i] * 1ll * let[i];
  for (int i = 0; i < 10; ++i) ans += num[i] * 1ll * num[i];
  cout << ans;
  return 0;
}