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.*;

public class test
{

	public static void main(String[] args)
	{
		new test().run();
	}

	long[] dp1 = null;
	long[] dp2 = null;
	long[] dp3 = null;

	boolean isPalin(char[] s, int start, int len)
	{
		for (int i = 0; i < len / 2; i++)
		{
			if (s[start + i] != s[start + len - 1 - i])
				return false;
		}

		return true;
	}

	void run()
	{
		Scanner in = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);

		char[] s = in.next().toCharArray();
		dp1 = new long[s.length];
		dp2 = new long[s.length];
		dp3 = new long[s.length];

		for (int i = 0; i < s.length; i++)
		{
			int x = i;
			int y = i;

			while (x >= 0 && y < s.length && s[x] == s[y])
			{
				dp1[x]++;
				dp2[y]++;

				x--;
				y++;
			}

			if (i < s.length - 1 && s[i] == s[i + 1])
			{
				x = i;
				y = i + 1;

				while (x >= 0 && y < s.length && s[x] == s[y])
				{
					dp1[x]++;
					dp2[y]++;

					x--;
					y++;
				}
			}

		}

		long tt = 0;
		for (int i = s.length - 1; i >= 0; i--)
		{
			tt += dp1[i];
			dp3[i] = tt;
		}

		long total = 0;
		for (int i = 0; i < s.length - 1; i++)
		{
			total += dp2[i] * dp3[i + 1];
		}

		out.println(total);

		out.close();
	}
}