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.math.BigDecimal;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;

public class A {

    static int max = 100000;

    // compare class //
    static class compare {
        int a;
        int b;

        public compare(int a, int b) {
            this.a = a;
            this.b = b;
        }
    }

    // comparator class //
    static Comparator<compare> cmp = new Comparator<compare>() {
        public int compare(compare a, compare b) {
            if (a.a > b.b)
                return 1;
            else if (a.a < b.b)
                return -1;
            else
                return 0;
        }
    };

    static int getSum(long a) {

        int result = 0;
        while (a > 0) {
            result += a % 10;
            a /= 10;
        }

        return result;
    }

    static long ipow(long base, long exp) {
        long result = 1;
        for (int i = 0; i < exp; i++)
            result *= base;

        return result;
    }

    public static void main(String[] args) {

        Scanner s = new Scanner(System.in);

        String str = s.nextLine();
        int[] chr = new int[128];

        for (int i = 0; i < str.length(); i++)
            chr[str.charAt(i)]++;

        long result = 0;

        for (int a : chr)
            result += (ipow(a, 2));

        System.out.println(result);
    }
}