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 com.sun.media.sound.FFT;

import java.io.*;
import java.text.NumberFormat;
import java.util.*;

/**
 * Created by VIET ANH on 27/04/2015.
 */
public class test {

    private static int T;
    private static long oo=Long.MAX_VALUE;
    private static long base=1000000007;

    public static void main(String[] args) throws IOException {
        Scanner in = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        String tmp = in.next();
        StringBuilder s = new StringBuilder();
        for (int i = 0; i < tmp.length(); i++) {
            s.append(tmp.charAt(i));
            s.append('@');
        }
        int[] cnt = new int[s.length()];
        boolean[][] f = new boolean[s.length()][s.length()];
        long res = 0;
        for (int i = 0; i < s.length(); i++) {
            int j=i+1;
            f[i][i]=true;
            while (j < s.length() && 2 * i - j >= 0 && s.charAt(j) == s.charAt(2 * i - j)) {
                f[j][2*i-j]=true;
                j++;
            }
        }
        for (int i = 0; i < s.length(); i++) {
            if (i>0) cnt[i]=cnt[i-1];
            if (i%2==0)
            for (int j = 0; j <= i; j++) {
                if (f[i][j]) {
                    cnt[i]++;
                    if (j>0) res+=cnt[j-1];
                }
            }
        }
        out.print(res);
        out.close();
    }
}