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

public class Solution {
	static final String FileName = "test";
	
	void solve() throws IOException {
		char[] s = br.readLine().toCharArray();
		int[] cnt = new int[256];
		for (char c:s) cnt[(int)c]++;
		long res = 0;
		for (int x:cnt) res += x*1l*x;
		out.println(res);
	}
	
	
	
	
	
	
	
	
	
	
	
	
	
	
    static final int inf = 1000000000+10;
    static final long linf = 1l<<62;
    static final double eps = 1e-8;
    BufferedReader br;
    StringTokenizer st;
    PrintWriter out;    
    
    
    public void run() {
        try {
            //br = new BufferedReader(new FileReader(FileName+".in"));
            //out = new PrintWriter(FileName+".out");
            br = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
            //br = new BufferedReader(new FileReader("input.txt"));
            //out = new PrintWriter("output.txt");
            solve();
            br.close();
            out.close();
        } catch (IOException e) {
            e.printStackTrace();
            System.exit(1);
        }
    }
    String next() throws IOException {
        while (st == null || !st.hasMoreTokens()) {
            String s = br.readLine();
            if (s == null)
                    return null;
            st = new StringTokenizer(s);
        }
        return st.nextToken();
    }
    double nextDouble() throws IOException {
       return Double.parseDouble(next());
    }

    int nextInt() throws IOException {
            return Integer.parseInt(next());
    }
    long nextLong() throws IOException {
            return Long.parseLong(next());
    }
    public static void main(String[] args) {        
    	new Solution().run();    
    }    
}