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.BufferedInputStream;
import java.util.Arrays;

public class C050B {

	public void solve() throws Exception {
		String s = nextWord();
		long res = 0;
		char[] arr = s.toCharArray();
		Arrays.sort(arr);
		char akt = arr[0];
		long count = 1;
		for (int i = 1; i < arr.length; i++) {
			if (akt == arr[i]) {
				count++;
			} else {
				akt = arr[i];
				res += count * count;
				count = 1;
			}
		}
		res += count * count;
		println(res);
	}

	// ------------------------------------------------------

	void debug(Object... os) {
		System.err.println(Arrays.deepToString(os));
	}

	void print(Object... os) {
		if (os != null && os.length > 0)
			System.out.print(os[0].toString());
		for (int i = 1; i < os.length; ++i)
			System.out.print(" " + os[i].toString());
	}

	void println(Object... os) {
		print(os);
		System.out.println();
	}

	BufferedInputStream bis = new BufferedInputStream(System.in);

	String nextWord() throws Exception {
		char c = (char) bis.read();
		while (c <= ' ')
			c = (char) bis.read();
		StringBuilder sb = new StringBuilder();
		while (c > ' ') {
			sb.append(c);
			c = (char) bis.read();
		}
		return new String(sb);
	}

	String nextLine() throws Exception {
		char c = (char) bis.read();
		while (c <= ' ')
			c = (char) bis.read();
		StringBuilder sb = new StringBuilder();
		while (c != '\n' && c != '\r') {
			sb.append(c);
			c = (char) bis.read();
		}
		return new String(sb);
	}

	int nextInt() throws Exception {
		return Integer.parseInt(nextWord());
	}

	long nextLong() throws Exception {
		return Long.parseLong(nextWord());
	}
	
	public static void main(String[] args) throws Exception {
		new C050B().solve();
	}
}