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.math.BigInteger;
import java.util.*;
import static java.lang.Math.*;

public class Solution {

    BufferedReader in;
    PrintWriter out;
    StringTokenizer st;

    void solve() throws IOException {
        char[] s = ns().toCharArray();
        int[] v = new int[1000];
        for (int i = 0; i < s.length; ++i)
            v[s[i]]++;
        BigInteger ret = BigInteger.ZERO;
        for (int i = 0; i < s.length; ++i) {
            ret = ret.add(BigInteger.valueOf(v[s[i]]));
        }
        out.println(ret);
    }

    public Solution() throws IOException {
        Locale.setDefault(Locale.US);
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        in.close();
        out.close();
    }

    String ns() throws IOException {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }

    int ni() throws IOException {
        return Integer.valueOf(ns());
    }

    long nl() throws IOException {
        return Long.valueOf(ns());
    }

    double nd() throws IOException {
        return Double.valueOf(ns());
    }

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