Problem: You are given a non-empty string s consisting of lowercase letters. Find the number of pairs of non-overlapping palindromic substrings of this string.

In a more formal way, you have to find the quantity of tuples (a, b, x, y) such that 1 ≤ a ≤ b < x ≤ y ≤ |s| and substrings s[a... b], s[x... y] are palindromes.

A palindrome is a string that can be read the same way from left to right and from right to left. For example, "abacaba", "z", "abba" are palindromes.

A substring s[i... j] (1 ≤ i ≤ j ≤ |s|) of string s = s1s2... s|s| is a string sisi + 1... sj. For example, substring s[2...4] of string s = "abacaba" equals "bac".

Input

The first line of input contains a non-empty string s which consists of lowercase letters ('a'...'z'), s contains at most 2000 characters.

Output

Output a single number — the quantity of pairs of non-overlapping palindromic substrings of s.

Please do not use the %lld format specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d format specifier.

Examples

Input

aa


Output

1


Input

aaa


Output

5


Input

abacaba


Output

36

Solution (in JAVA):

Solution:

import java.io.File;
import java.io.FileOutputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {

        try {
            Scanner sc = new Scanner(System.in);
            //Scanner sc = new Scanner(new File("input.txt"));

            ArrayList<Long> left = new ArrayList<Long>();
            ArrayList<Long> right = new ArrayList<Long>();

            String str = sc.next();

            for (int i = 0; i < str.length(); ++i) {
                left.add(0l);
                right.add(0l);
            }

            long res = 0;

            for (int i = 0; i < str.length(); ++i) {
                int size = 0;
                while (true) {
                    if ((i - size < 0) || (i + size >= str.length())) {
                        break;
                    }
                    if (str.charAt(i - size) != str.charAt(i + size)) {
                        break;
                    }

                    left.set(i + size, left.get(i + size) + 1);
                    right.set(i - size, right.get(i - size) + 1);

                    size++;
                }
            }

               for (int i = 0; i < str.length(); ++i) {
                int size = 0;
                while (true) {
                    if ((i - size < 0) || (i+1 + size >= str.length())) {
                        break;
                    }
                    if (str.charAt(i - size) != str.charAt(i+1 + size)) {
                        break;
                    }

                    left.set(i + 1 + size, left.get(i+1 + size) + 1);
                    right.set(i - size, right.get(i - size) + 1);

                    size++;
                }
            }


           



            for (int i = 0; i < str.length(); ++i) {
                long tmp = 0;
                for (int j = i + 1; j < str.length(); ++j) {
                    tmp += right.get(j);
                }

                res += left.get(i) * tmp;
            }

            System.out.print(res);

        } catch (Exception e) {
            e.printStackTrace();
        }

    }
}