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.*;
//import java.util.Arrays;
import java.math.BigInteger;


public class Main {
   
    
    public static BigInteger getPalCount(String s){
        int length = s.length();
        boolean isPal[][] = new boolean[length][length];
        int p = 0;
        while(p<length){
            for(int i = 0;i+p<length;i++){
                if(p==0){
                   isPal[i][i+p] = true; 
                }
                else if(p==1){
                    isPal[i][i+p] = s.charAt(i)==s.charAt(i+p); 
                }
                else{
                    if(s.charAt(i)==s.charAt(i+p))
                        isPal[i][i+p] = isPal[i+1][i+p-1]; 
                
                    else
                        isPal[i][i+p] = false; 
                }
            }
            p++;
        }
        int countLeft[] = new int[length];
        int countRight[] = new int[length];
        for(int i = 0; i<length;i++){
            for(int j = 0; j<=i;j++){
                
            if(isPal[j][i])
                countLeft[i]++;
                
            }
            //System.out.println(countLeft[i]+" "+i);
        }
        for(int i = 0; i<length;i++){
            for(int j = i; j<length;j++){
                
            if(isPal[i][j])
                countRight[i] += 1;
            }
            //System.out.println(countRight[i]+" "+i);
        }
        int sum = 0;
        for(int i = length-1;i>0;i--){
            sum += countRight[i];
        }
        BigInteger res = BigInteger.ZERO;
        for(int i = 0; i<length-1;i++){
            res = res.add(new BigInteger(String.valueOf(sum*countLeft[i])));
            sum -= countRight[i+1];
        }
        return res;
    }
    public static void main(String args[]) {
        try {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
            String temp  = br.readLine();
            System.out.println(getPalCount(temp));
            br.close();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
   }