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.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.StringTokenizer;


public class B50 {
	BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
	StringTokenizer st = null;
	
	
	private void solution() throws IOException {
		String s = nextToken();
		long a[] = new long[256];
		long ans = 0;
		for (int i = 0; i < s.length(); i ++) a[s.charAt(i)] ++;
		for (int i = 0; i < 256; i ++) ans += a[i]*a[i];
		System.out.print(ans);
	
	}

	String nextToken() throws IOException {
		if (st == null || !st.hasMoreTokens()) {
			st = new StringTokenizer(bf.readLine());
		}
		return st.nextToken();
	}
	int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}

	public static void main(String args[]) throws IOException {
		new B50().solution();
	}
	
}