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


public class CodeD 
{
	static class Scanner
	{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		
		public String nextLine()
		{
			try
			{
				return br.readLine();
			}
			catch(Exception e)
			{
				throw(new RuntimeException());
			}
		}
		
		public String next()
		{
			while(!st.hasMoreTokens())
			{
				String l = nextLine();
				if(l == null)
					return null;
				st = new StringTokenizer(l);
			}
			return st.nextToken();
		}
		
		public int nextInt()
		{
			return Integer.parseInt(next());
		}
		
		public long nextLong()
		{
			return Long.parseLong(next());
		}
		
		public double nextDouble()
		{
			return Double.parseDouble(next());
		}
		
		public int[] nextIntArray(int n)
		{
			int[] res = new int[n];
			for(int i = 0; i < res.length; i++)
				res[i] = nextInt();
			return res;
		}
		
		public long[] nextLongArray(int n)
		{
			long[] res = new long[n];
			for(int i = 0; i < res.length; i++)
				res[i] = nextLong();
			return res;
		}
		
		public double[] nextDoubleArray(int n)
		{
			double[] res = new double[n];
			for(int i = 0; i < res.length; i++)
				res[i] = nextDouble();
			return res;
		}
		public void sortIntArray(int[] array)
		{
			Integer[] vals = new Integer[array.length];
			for(int i = 0; i < array.length; i++)
				vals[i] = array[i];
			Arrays.sort(vals);
			for(int i = 0; i < array.length; i++)
				array[i] = vals[i];
		}
		
		public void sortLongArray(long[] array)
		{
			Long[] vals = new Long[array.length];
			for(int i = 0; i < array.length; i++)
				vals[i] = array[i];
			Arrays.sort(vals);
			for(int i = 0; i < array.length; i++)
				array[i] = vals[i];
		}
		
		public void sortDoubleArray(double[] array)
		{
			Double[] vals = new Double[array.length];
			for(int i = 0; i < array.length; i++)
				vals[i] = array[i];
			Arrays.sort(vals);
			for(int i = 0; i < array.length; i++)
				array[i] = vals[i];
		}

		public String[] nextStringArray(int n) 
		{	
			String[] vals = new String[n];
			for(int i = 0; i < n; i++)
				vals[i] = next();
			return vals;
		}
		
		Integer nextInteger()
		{
			String s = next();
			if(s == null)
				return null;
			return Integer.parseInt(s);
		}
		
		int[][] nextIntMatrix(int n, int m)
		{
			int[][] ans = new int[n][];
			for(int i = 0; i < n; i++)
				ans[i] = nextIntArray(m);
			return ans;
		}
	}
	
	public static void main(String[] args)
	{
		Scanner sc = new Scanner();
		long[] suma = new long[256];
		for(char c : sc.next().toCharArray()) suma[c]++;
		long cuenta = 0;
		for(int i = 0; i < suma.length; i++) cuenta += (suma[i] * suma[i]);
		System.out.println(cuenta);
	}

}