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.*;
import java.util.*;

public class B
{
PrintWriter out;
BufferedReader in;
StringTokenizer ss;

   void dbg(String s){System.out.println(s);};

   String next_token() throws IOException
   {while (!ss.hasMoreTokens())ss=new StringTokenizer(in.readLine());
    return ss.nextToken();}

   Double _double() throws IOException
   {return Double.parseDouble (next_token());}

   int _int() throws IOException
   {return Integer.parseInt (next_token());}

   long _long() throws IOException
   {return Long.parseLong (next_token());}

void run()throws IOException
{

in = new BufferedReader(new InputStreamReader(System.in));
out= new PrintWriter(System.out);

String name="";
/*in = new BufferedReader(new FileReader(name+".in"));
out = new PrintWriter(new File(name+".out"));  */
ss = new StringTokenizer("  ");
String S=next_token();
long A[]=new long[512];
    for(int i=0;i<S.length();i++)A[S.charAt(i)]++;
    long ans=0;
    for(int i=0;i<512;i++){
        ans+=A[i]*A[i];
    }
out.println(ans);
out.close();      
}

public static void main(String[] args)throws Exception
{
try{new B().run();}catch (Exception e){System.out.println("Runtime Error");};
}
}