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.math.BigInteger;
import java.util.StringTokenizer;

public class B implements Runnable {

    // leave empty to read from stdin/stdout
    private static final String TASK_NAME_FOR_IO = "";

    // file names
    private static final String FILE_IN = TASK_NAME_FOR_IO + ".in";
    private static final String FILE_OUT = TASK_NAME_FOR_IO + ".out";

    BufferedReader in;
    PrintWriter out;
    StringTokenizer tokenizer = new StringTokenizer("");

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

    char[] c;
    long[] cnt = new long[1000];

    private void solve() throws IOException {
        c = nextToken().toCharArray();
        for (char v : c) {
            cnt[v]++;
        }

        long result = 0;
        for (long l : cnt) {
            result += l * l;
        }
        out.print(result);
    }

    public void run() {
        long timeStart = System.currentTimeMillis();
        try {

            if (TASK_NAME_FOR_IO.length() > 0) {
                in = new BufferedReader(new FileReader(FILE_IN));
                out = new PrintWriter(new FileWriter(FILE_OUT));
            } else {
                in = new BufferedReader(new InputStreamReader(System.in));
                out = new PrintWriter(new OutputStreamWriter(System.out));
            }

            solve();

            in.close();
            out.close();
        } catch (IOException e) {
            throw new IllegalStateException(e);
        }
        long timeEnd = System.currentTimeMillis();
        System.out.println("Time spent: " + (timeEnd - timeStart) + " ms");
    }

    private String nextToken() throws IOException {
        while (!tokenizer.hasMoreTokens()) {
            String line = in.readLine();
            if (line == null) {
                return null;
            }
            tokenizer = new StringTokenizer(line);
        }
        return tokenizer.nextToken();
    }

    private int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    private BigInteger nextBigInt() throws IOException {
        return new BigInteger(nextToken());
    }

    private long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    private double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

}