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 java.io.BufferedReader;
import java.io.IOException;

import java.io.InputStream;
import java.io.InputStreamReader;
import static java.lang.Integer.parseInt;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.util.regex.Matcher;
import java.util.regex.Pattern;



 public class NewClass {

     
     static class escaner
       {
           BufferedReader in;
           StringTokenizer st;

        public escaner() {
       
            in=new BufferedReader(new InputStreamReader(System.in));
            st=new StringTokenizer("");
            }
        
        public int nextInt() throws IOException
        {
            if(st.hasMoreTokens())return Integer.parseInt(st.nextToken());
            else
            { st=new StringTokenizer(in.readLine());
            return nextInt();
            
            }
            
        }
             public String next() throws IOException
        {
            if(st.hasMoreTokens())return st.nextToken();
            else
            { st=new StringTokenizer(in.readLine());
            return next();
            
            }
            
        }
                public String nextLine() throws IOException
        {
            return in.readLine();
           
            
        
       }
     }
  
    public static void main(String[] args) throws Throwable {

    
       
        
	escaner sc=new escaner();
	
        String s=sc.next();
        char[]k=s.toCharArray();
        long n=s.length();
    
        int l=s.length();
        
        Map<Character,Long> m=new HashMap<Character,Long>();
        
        
        for (int i = 0; i < l; i++) {
            
            if(m.containsKey(k[i]))
            {
                long t=m.get(k[i])+1;
                m.put(k[i], t);
                        }
            else
                 m.put(k[i], (long)1);
            
        }
        for (Map.Entry<Character, Long> entry : m.entrySet()) {
            
            long x=entry.getValue()-1;
            if(x%2==0)
            
            n+= ((x*(x/2))+x/2)*2;
            else
            {
                x--;
                long mm=((x*(x/2))+x/2);
              
                mm+=x+1;
                  n+= mm*2;
            }
            
        }
        

        System.out.println(n);

    }
}