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.StringTokenizer;

public class Q50B implements Runnable{
    private void solve() throws IOException {
        char[] s = nextToken().toCharArray();
        int[]   ap=  new int['z'-'a'+1+'9'-'0'+1];
        long count=s.length;
        for (int i = 0; i < s.length; i++) {
            char c = s[i];
            int index = c>='0'&&c<='9'?c-'0':c-'a'+10;
            count+=ap[index]*2;
            ap[index]++;
        }
        writer.print(count);
    }

    public static void main(String[] args) {
        new Q50B().run();
    }

    BufferedReader reader;
    StringTokenizer tokenizer;
    PrintWriter writer;

    public void run() {
        try {

            boolean oj = System.getProperty("ONLINE_JUDGE") != null;

            reader = oj ? new BufferedReader(new InputStreamReader(System.in)) :
                    new BufferedReader(new FileReader(new File("input.txt")));


            tokenizer = null;
            writer = new PrintWriter(System.out);
            solve();
            reader.close();
            writer.close();
        } catch (Exception e) {
            e.printStackTrace();
            System.exit(1);
        }
    }

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

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

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

    String nextToken() throws IOException {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            tokenizer = new StringTokenizer(reader.readLine());
        }
        return tokenizer.nextToken();
    }
}