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 JAVA):

Solution:


import javax.print.DocFlavor;
import javax.swing.*;
import javax.swing.text.StyleContext;
import java.awt.image.AreaAveragingScaleFilter;

import javax.print.attribute.Attribute;
import javax.print.attribute.HashAttributeSet;
import java.io.*;
import java.lang.reflect.Array;
import java.math.BigInteger;
import java.nio.channels.AcceptPendingException;
import java.nio.charset.IllegalCharsetNameException;
import java.util.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.util.Collections;
import java.io.InputStreamReader;
import java.util.logging.SimpleFormatter;


    public class Main1 {





        public static void main(String[] args) {




                FastScanner in = new FastScanner();
                PrintWriter out = new PrintWriter(System.out);

                char a[] = in.next().toCharArray() ;
                Arrays.sort(a);
                int ct = 1 ;
                long temp = 0 ;
            for (int i = 1; i <a.length ; i++) {
                if (a[i] == a[i-1])
                    ct++ ;
                else {
                    temp += (long) ct *ct ;
                    ct = 1 ;
                }
            }

            System.out.println(temp + (long) ct *ct);

















                out.flush();
                out.close();




        }


















        static class FastScanner {
            BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
            StringTokenizer st = new StringTokenizer("");

            String next() {
                while (!st.hasMoreTokens())
                    try {
                        st = new StringTokenizer(br.readLine());
                    } catch (IOException e) {
                        e.printStackTrace();
                    }
                return st.nextToken();
            }

            int nextInt() {
                return Integer.parseInt(next());
            }

            int[] readArray(int n) {
                int[] a = new int[n];
                for (int i = 0; i < n; i++) a[i] = nextInt();
                return a;
            }


            long nextLong() {
                return Long.parseLong(next());
            }
        }


    }