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 MAXN = (long long)((1e5) + 100);
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long max(long long a, long long b) {
  if (a > b)
    return a;
  else
    return b;
}
long long min(long long a, long long b) {
  if (a < b)
    return a;
  else
    return b;
}
void pr(long long x) { cout << x; }
void prl(long long x) { cout << x << "\n"; }
long long cbrt(long long x) {
  long long lo = 1, hi = min(2000000ll, x);
  while (hi - lo > 1) {
    long long mid = (lo + hi) / 2;
    if (mid * mid * mid < x) {
      lo = mid;
    } else
      hi = mid;
  }
  if (hi * hi * hi <= x)
    return hi;
  else
    return lo;
}
const long long dx[4] = {-1, 1, 0, 0};
const long long dy[4] = {0, 0, -1, 1};
long long XX[] = {-1, -1, -1, 0, 0, 1, 1, 1};
long long YY[] = {-1, 0, 1, -1, 1, -1, 0, 1};
const long long N = (long long)(2 * 1e5 + 10);
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  long long T = 1;
  while (T--) {
    string s;
    cin >> s;
    unordered_map<char, long long> mpp;
    for (auto it : s) mpp[it]++;
    long long cnt = 0;
    for (auto it : s) {
      cnt += mpp[it];
    }
    cout << cnt;
  }
}