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.IOException;
import java.io.PrintWriter;
import java.util.Scanner;

public class B47 implements Runnable {

    public Scanner in;

    public PrintWriter out;

    B47() throws IOException {
        in = new Scanner( System.in );
        // in = new StreamTokenizer( new InputStreamReader( System.in ) );
        out = new PrintWriter( System.out );
    }

    void close() throws IOException {
        out.close();
    }

    public void run() {
        try {
            solve();
            close();
        } catch ( Exception e ) {
            e.printStackTrace();
        }
    }

    public void solve() throws IOException {
        char[] c = in.next().toCharArray();
        long[] l = new long['z' - 'a' + 1];
        long[] d = new long[10];
        for ( int i = 0; i < c.length; i ++ ) {
            if ( c[i] >= 'a' && c[i] <= 'z' ) {
                l[c[i] - 'a'] ++;
            } else {
                d[c[i] - '0'] ++;
            }
        }
        long r = 0;
        for ( int i = 0; i < l.length; i ++ ) {
            r += l[i] * l[i];
        }
        for ( int i = 0; i < d.length; i ++ ) {
            r += d[i] * d[i];
        }
        out.println( r );
    }

    public static void main( String[] args ) throws IOException {
        new Thread( new B47() ).start();
    }
}