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:
/** Team Proof
Java template
*/

import java.io.*;
import java.util.*;
import java.math.*;

public class cf
{
	BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
	PrintWriter out = new PrintWriter (System.out);
	
	void debug(Object...os)
	{
		out.println(Arrays.deepToString(os));
	}
	
// 	int ni() throws IOException
// 	{
// 		in.nextToken();
// 		return (int)in.nval;
// 	}
// 	
// 	long nl() throws IOException
// 	{
// 		in.nextToken();
// 		return (long)in.nval;
// 	}
// 	
// 	double nd() throws IOException
// 	{
// 		in.nextToken();
// 		return in.nval;
// 	}
// 	
// 	String ns() throws IOException
// 	{
// 		in.nextToken();
// 		return in.sval;
// 	}
// 	
	void execute() throws IOException
	{
		totalCases = 1;
		for(testNum = 1; testNum <= totalCases; testNum++)
		{
			if(!input())
				break;
			solve();
			out.flush();
		}
	}
	
	int totalCases, testNum;
	
	void solve()
	{
	}
	
	boolean input() throws IOException
	{
		String s= in.readLine();
		int [] digs = new int [300];
		for(int i = 0; i < s.length(); i++)
			digs[s.charAt(i)] ++;
		long ans = 0;
		for(int a : digs )
		{
			ans += ( long ) a * ( long ) a;
		}
		
		System.out.println(ans);
		
		return true;
	}
	
	public static void main(String[] args) throws IOException
	{
		new cf().execute();
	}
}