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:

/*
 * To change this template, choose Tools | Templates
 * and open the template in the editor.
 */


import java.util.Scanner;

/**
 *
 * @author JAVA
 */
public class d {

    static String reverse(String s){
        String str = "";
        for (int i = s.length()-1; i>=0; i--)
            str += s.charAt(i);
        return str;
    }
    
    static boolean isPalindrom(String s){
        if (s.length()%2 == 0){
            //System.out.println(s.substring(0, s.length()/2) + " " + s.substring((s.length()/2), s.length()));
            
            if (s.substring(0, s.length()/2).equals(reverse(s.substring((s.length()/2), s.length()))))
                return true;
        }else {
        //System.out.println(s.substring(0, s.length()/2) + " " + reverse(s.substring((s.length()/2)+1, s.length())));
        if (s.substring(0, s.length()/2).equals(reverse(s.substring((s.length()/2)+1, s.length()))))
                return true;
        }
        return false;
    }
    
    
    
    public static void main(String[] args) {
        Scanner in = new Scanner (System.in);
        String s = in.next();
        //System.out.println(s.hashCode());
        //System.out.println(isPalindrom(s));
        boolean t[][] = new boolean[s.length()][s.length()];
        long d[] = new long[s.length()+1];
        
        for (int i =0; i<s.length(); i++){
            int l = i; int r = i;
            while (l>=0 && r<s.length()){
                if (s.charAt(l) == s.charAt(r))
                {
                    if (!t[l][r]){
                        t[l][r] = true;
                        d[l]++;
                    }
                    l--; r++;
                }else break;
            }
            l = i; r =i+1;
            while (l>=0 && r<s.length()){
                if (s.charAt(l) == s.charAt(r))
                {
                    if (!t[l][r])
                    {
                        t[l][r] = true;
                        d[l]++;
                    }
                    l--; r++;
                }else break;
            }
        }
        int sum = 0;
        int sum1[] = new int[s.length()+1];
        for (int i = s.length()-2; i>=0; i--){
            d[i] += d[i+1];
        }
        /*
        for (int i=0; i<s.length(); i++)
            System.out.print(d[i] + " ");
        System.out.println("---");*/
        long count = 0;
        for (int i =0; i<s.length(); i++){
            for (int j = 0; j<s.length(); j++)
                if (t[i][j]){
          //          System.out.print(d[j+1] + " ");
                    count += d[j+1];
                }
        }
        System.out.println(count);
    }
}