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.util.Scanner;

/**
 * 
 */

/**
 * created Mar 9, 2012
 * 
 * @author dkorduban
 */
public class D implements Runnable {

    private char[] s;
    private boolean[][] was;
    private boolean[][] sav;

    /**
     * @param args
     */
    public static void main(final String[] args) {
        // TODO Auto-generated method stub
        new D().run();
    }

    /*
     * (non-Javadoc)
     * @see java.lang.Runnable#run()
     */
    @Override
    public void run() {
        final Scanner in = new Scanner(System.in);
        s = in.next().toCharArray();
        final int[] palRight = new int[s.length];
        sav = new boolean[s.length][s.length];
        was = new boolean[s.length][s.length];
        for (int i = 0; i < s.length; ++i) {
            for (int j = i; j < s.length; ++j) {
                if (isPalindrom(i, j)) {
                    ++palRight[i];
                }
            }
        }

        final int[] sumRight = new int[s.length];
        sumRight[s.length - 1] = palRight[s.length - 1];
        for (int i = s.length - 2; i >= 0; --i) {
            sumRight[i] = sumRight[i + 1] + palRight[i];
        }
        long ans = 0;
        for (int i = 0; i < s.length; ++i) {
            for (int j = i; j < s.length - 1; ++j) {
                if (isPalindrom(i, j)) {
                    ans += sumRight[j + 1];
                }
            }
        }

        System.out.println(ans);
    }

    /**
     * @param i
     * @param j
     * @return
     */
    private boolean isPalindrom(final int i, final int j) {
        if (i == j) {
            return true;
        }
        if (i + 1 == j) {
            return s[i] == s[j];
        }
        if (was[i][j]) {
            return sav[i][j];
        }
        was[i][j] = true;
        return sav[i][j] = (s[i] == s[j]) && isPalindrom(i + 1, j - 1);
    }

}