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.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) throws IOException {
		
		Scanner sc = new Scanner(System.in);
		
		String entrada;
		
		char[] caracteres;
	    long[]  letras = new long[26];
	    long[]  numeros = new long[10];
	    long numPairs = 0;
	 
	    entrada = sc.nextLine();
	    caracteres = entrada.toCharArray();
	 
	    for(int i = 0; i < caracteres.length; i++){
	        if(((int)caracteres[i] >= 97) && ((int)caracteres[i] <= 122)){
	            letras[(int)caracteres[i] - 97]++;
	        }
	        else if(((int)caracteres[i] >= 48) && ((int)caracteres[i] <= 57)){
	            numeros[(int)caracteres[i] - 48]++;
	        }
	    }
	 
	    for(int i = 0; i < 10; i++){
	        if(numeros[i] != 0){
	            numPairs += Math.pow(numeros[i], 2);
	        }
	        if(letras[i] != 0){
	            numPairs += Math.pow(letras[i], 2);
	        }
	    }
	    for(int i = 10; i < 26; i++){
	        if(letras[i] != 0){
	            numPairs += Math.pow(letras[i], 2);
	        }
	    }
	    System.out.println(numPairs);
	 
	}
}