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.awt.*;
import java.awt.event.*;
import java.awt.geom.*;
import java.io.*;
import java.math.*;
import java.text.*;
import java.util.*;

public class Main {
	static BufferedReader br;
	static StringTokenizer st;
	static PrintWriter pw;

	public static void main(String[] args) throws IOException	{
		br = new BufferedReader(new InputStreamReader(System.in));
		pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		String next = nextToken();
		int n = next.length();
		int[] begin = new int[n];
		int[] end = new int[n];
		outer: for(int i = 0; i < next.length(); i++)	{
			begin[i]++;
			end[i]++;
			int diff = 1;
			while(true)	{
				if(i-diff < 0 || i+diff >= next.length())
					continue outer;
				if(next.charAt(i-diff) != next.charAt(i+diff))
					continue outer;
				begin[i-diff]++;
				end[i+diff]++;
				diff++;
			}
		}
		outer: for(int i = 0; i + 1 < next.length(); i++)	{
			if(next.charAt(i) != next.charAt(i+1))
				continue outer;
			begin[i]++;
			end[i+1]++;
			int diff = 1;
			while(true)	{
				if(i-diff < 0 || i+1+diff >= next.length())
					continue outer;
				if(next.charAt(i-diff) != next.charAt(i+1+diff))
					continue outer;
				begin[i-diff]++;
				end[i+1+diff]++;
				diff++;
			}
		}
		long ret = 0;
		for(int i = 0; i < n; i++)	{
			for(int j = i+1; j < n; j++)	{
				ret += end[i]*1L*begin[j];
			}
		}
		pw.println(ret);
		pw.close();
	}

	public static long readLong() throws IOException	{
		return Long.parseLong(nextToken());
	}
	public static double readDouble() throws IOException	{
		return Double.parseDouble(nextToken());
	}
	public static int readInt() throws IOException	{
		return Integer.parseInt(nextToken());
	}
	public static short readShort() throws IOException	{
		return Short.parseShort(nextToken());
	}
	public static String nextToken() throws IOException	{
		while(st == null || !st.hasMoreTokens())	{
			if(!br.ready())	{
				pw.close();
				System.exit(0);
			}
			st = new StringTokenizer(br.readLine());
		}
		return st.nextToken();
	}
}