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 by slycer on 9/17/15.
 */
public class Palindromepairs {

    public static void main ( String [] arg ){
        Scanner s = new Scanner ( System.in );

        String line = s.nextLine();
        boolean [][] isPalindrome = new boolean[line.length()][line.length()];

        for ( int i=0; i<line.length(); i++ ){
            isPalindrome[i][i] = true;
        }


        for ( int j=1; j<line.length(); j++ ){
            for ( int i=0; i+j<line.length(); i++ ){

                if ( line.charAt(i) == line.charAt( i+j ) ){
                    if ( j == 1 ){
                        isPalindrome[i][i+j] = true;
                    } else {
                        isPalindrome[i][i+j] = isPalindrome[i+1][i+j-1];
                    }
                }
               // System.out.println( i + " " + (i+j) + " " + isPalindrome[i][i+j]);
            }
        }

        // numarul de palindroame distincte intre [i,j]
        int [][] palindromeCount = new int[ line.length() + 1 ][ line.length() + 1  ];

        for ( int i=0; i<line.length(); i++ ){
            palindromeCount[i][i] ++;
        }
        for ( int j=1; j<line.length(); j++ ){
            for ( int i=0; i+j<line.length(); i++ ){
                if ( isPalindrome[i][i+j] ){
                    palindromeCount[i][i+j] = 1;
                }
                palindromeCount[i][i+j] += palindromeCount[i+1][i+j] + palindromeCount[i][i+j-1] - palindromeCount[i+1][i+j-1];
                //System.out.println( i + " " + (i + j) + " :" + palindromeCount[i][i+j] );
            }
        }

        // numaram combinatiile
        long combinatii = 0;
        for ( int i=0; i<line.length(); i++ ){
            for ( int j=i; j<line.length(); j++ ){
                if ( isPalindrome[i][j] ){
                    combinatii += palindromeCount[j+1][line.length()-1];
                }
            }
        }

        System.out.println( combinatii );

    }

}