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.InputStreamReader;

public class Main {

	// split de array
	public static int[ ] atoi( String cad ) {
		String read[] = cad.split( " " );
		int res[] = new int[ read.length ];
		for ( int i = 0; i < read.length; i++ ) {
			res[ i ] = Integer.parseInt( read[ i ] );
		}
		return res;
	}

	public static void main( String[ ] args ) throws IOException {
		BufferedReader in = new BufferedReader( new InputStreamReader( System.in ) );
		char cad[] = in.readLine( ).toCharArray( );
		long arr[] = new long[ 27 ];
		long digits[] = new long[ 10 ];
		for ( int i = 0; i < cad.length; i++ ) {
			if ( cad[ i ] >= '0' && cad[ i ] <= '9' ) {
				digits[ cad[ i ] - '0' ]++;
			} else {
				arr[ cad[ i ] - 'a' ]++;
			}
		}
		long sum = 0;
		for ( int i = 0; i < arr.length; i++ ) {
			sum += arr[ i ] * arr[ i ];
		}
		for ( int i = 0; i < digits.length; i++ ) {
			sum += digits[ i ] * digits[ i ];
		}
		System.out.println( sum );

	}

}