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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class CF50B {
	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		String s = sc.nextLine();
		int f1[] = new int[26], f2[] = new int[10];
		for (int i = 0; i < s.length(); i++) {
			char c = s.charAt(i);
			if (Character.isAlphabetic(c))
				f1[c - 'a']++;
			else
				f2[c - '0']++;
		}

		long ans = 0;
		for (int i = 0; i < 26; i++) {
			ans += 1l * f1[i] * f1[i];
		}

		for (int i = 0; i < 10; i++) {
			ans += 1l * f2[i] * f2[i];
		}
		System.out.println(ans);
	}

	static class Scanner {
		BufferedReader br;
		StringTokenizer st;

		Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());

			return st.nextToken();
		}

		String nextLine() throws IOException {
			return br.readLine();
		}

		int nextInt() throws NumberFormatException, IOException {
			return Integer.parseInt(next());
		}

		boolean ready() throws IOException {
			return br.ready();
		}

	}
}