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.util.Scanner;

public class D {

	/**
	 * @param args
	 */
	static char[] s;
	static int[][] p;

	public static void main(String[] args)
	{
		Scanner scan = new Scanner(System.in);
		s = scan.next().toCharArray();
		
		long[] dp = new long[s.length+1];
		p = new int[s.length][s.length];
		
		for (int i1 = s.length - 1; i1 >= 0; i1--)
		{
			dp[i1] = dp[i1+1];
			for (int i2 = i1; i2 < s.length; i2++)
				if (s[i1] == s[i2] && palindrome(i1, i2))
				{
					dp[i1]++;
				}
		}
		
		long ret = 0;
		for(int i1 = 0 ;i1 < s.length;i1++)
			for (int i2 = i1 ; i2 < s.length-1; i2++)
				if (s[i1] == s[i2] && palindrome(i1, i2))
				{
					ret += dp[i2+1];
				}
		System.out.println(ret);
	}
	
	private static boolean palindrome(int f, int e)
	{
		if (p[f][e] != 0)
		{
			if (p[f][e] == 1)
				return true;
			else
				return false;
		}
		int n = (e - f + 1) / 2;
		for (int i = 0; i < n; i++)
			if (s[f + i] != s[e - i])
			{
				p[f][e] = -1;
				return false;
			}
		p[f][e] = 1;
		return true;
	}

}