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 Main {
    private void solve() throws IOException {
        String s = nextToken();
        long[] z = new long[1024];
        for (int i = 0; i < s.length(); i++) {
            z[s.charAt(i)]++;
        }
        long result = 0;
        for (int i = 0; i < 1024; i++) {
            result += z[i] * z[i];
        }
        System.out.println(result);
    }

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

    BufferedReader reader;
    StringTokenizer tokenizer;
    PrintWriter writer;

    public void run(String[] args) {
        try {
            if (args.length > 0 && args[0].equals("input.txt")) {
                reader = new BufferedReader(new FileReader(new File(args[0])));
            } else {
                reader = new BufferedReader(new InputStreamReader(System.in));
            }

            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();
    }
}