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.lang.Math;
import java.util.Scanner;
import java.io.*;
import java.util.*;

public class Some {
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        String s = in.readLine();
        int[] left = new int[s.length()];
        int[] right = new int[s.length()];
        Arrays.fill(left, 0);
        Arrays.fill(right, 0);
        for(int i = 0; i < s.length(); i++) { //odd palindromes
            for(int diff = 0; i - diff >= 0 && i + diff < s.length(); diff++) {
                int l = i - diff;
                int r = i + diff;
                if(s.charAt(l) == s.charAt(r)) {
                    //if ((i == 1) && (diff == 1)) System.out.println(left[i]);
                    left[r]++;
                    right[l]++;
                }
                else
                    break;
            }

        }
        
       
        for(int i = 0; i < s.length(); i++) { //even
            for(int diff = 0; i - diff >= 0 && i + diff + 1 < s.length(); diff++) {
                int l = i - diff;
                int r = i + diff + 1;
                if(s.charAt(l) == s.charAt(r)) {
                    left[r]++;
                    right[l]++;
                }
                else
                    break;
            }
        }/*
        for (int i = 0; i < s.length(); i++){
            System.out.print(left[i] + " ");
        }
        System.out.println();
        for (int i = 0; i < s.length(); i++){
            System.out.print(right[i] + " ");
        }*/
        long res = 0;
        for(int i = 0; i < s.length(); i++) {
            for(int j = i + 1; j < s.length(); j++) {
                res += left[i] * right[j];
            }
        }
        System.out.println(res);
    }
}