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.util.*;
import java.io.*;
import static java.util.Arrays.*;
import static java.lang.Math.*;
import java.math.*;

public class Main {

    void run() throws IOException {
        char[] s = token().toCharArray();
        int n = s.length;
        long[] v = new long[36];
        for (int i = 0; i < n; i++) {
            if (Character.isDigit(s[i])) {
                v[s[i] - '0' + 26]++;
            } else {
                v[s[i] - 'a']++;
            }
        }
        long r = 0;
        for (int i = 0; i < 36; i++) {
            r += v[i] * v[i];
        }
        System.out.println(r);
    }

    public static void main(String[] args) throws IOException {
        Locale.setDefault(Locale.US);
        final String FILENAME = "";
        //in = new BufferedReader(new FileReader(new File(FILENAME + ".in")));
        //out = new PrintWriter(new File(FILENAME + ".out"));
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        st = new StringTokenizer(" ");
        new Main().run();
        out.close();
    }

    static BufferedReader in;
    static PrintWriter out;
    static StringTokenizer st;

    String token() throws IOException {
        while (!st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }

    int nint() throws IOException {
        return Integer.parseInt(token());
    }

    long nlong() throws IOException {
        return Long.parseLong(token());
    }

    double ndouble() throws IOException {
        return Double.parseDouble(token());
    }
}