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.math.BigInteger;


public class Main {
    
  
    public static void main(String args[]) {
        try {
        
        //BufferedReader br = new BufferedReader(new FileReader(new File("input.txt")));
         BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        //
        
        String S = br.readLine();
        int count = 0;
        int l = S.length();
        boolean A[][] = new boolean[l][l];
        int p = 0;
        
        while(p<l){
            for(int i = 0;i+p<l;i++){
                if(p==0){
                    A[i][i+p] = true;
                }
                else if(p==1){
                    A[i][i+p] = (S.charAt(i)==S.charAt(i+p));
                }
                else{
                    A[i][i+p] = (S.charAt(i)==S.charAt(i+p)&&A[i+1][i+p-1]);
                }
            }
            p++;
        }
        
        int countArr[] = new int[l];
        count = 0;
        for(int i = l-1;i>=0;i--){
            
            for(int j = i;j<l;j++)
                if(A[i][j])
                    count++;
            countArr[i] = count;
        }
        BigInteger count1 = BigInteger.ZERO;
        
        for(int i = l-2;i>=0;i--){
            
            for(int j=i;j<=l-2;j++){
                
                if(A[i][j]==true){
                    count1 = count1.add(new BigInteger(String.valueOf(countArr[j+1]))); 
                    //System.out.println(count+" "+countArr[j+1]);
                }
            }
        }
        
        System.out.println(count1);
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
   }