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 PalindromePairs 
{

	public static void main(String[] args) 
	{
		Scanner sc = new Scanner(System.in);
		
		String s = sc.next();
		int n = s.length();
		
		boolean[][] pal = new boolean[n][n];
		long[][] dp = new long[n][n]; //num of substring palindromes within this range
		
		for(int i = 0; i < n; i++)
		{
			pal[i][i] = true;
			dp[i][i] = 1;
		}
		
		//dp palindromes
		for(int sz = 2; sz <= n; sz++)
		{
			for(int l = 0; l + sz - 1 < n; l++)
			{
				int r = l + sz - 1;
				
				if(s.charAt(l) == s.charAt(r) && (r == l+1 || pal[l+1][r-1]))
				{
					dp[l][r] += 1; //this itself is a palindrome
					pal[l][r] = true;
				}
					
				if(r == l + 1)
				{
					dp[l][r] += 2; //leftest and rightest by themself
				}
				else
				{
					dp[l][r] += - dp[l+1][r-1] + dp[l+1][r] + dp[l][r-1];
//					System.out.println("DUN " + l + " " + r + " => " + dp[l][r]);
//					System.out.println(-dp[l+1][r-1] + " " + dp[l+1][r] + " " + dp[l][r-1]);
					
//						dp[l][r] = 2 + dp[l+1][r-1] + dp[l+1][r] + dp[l][r-1];
				}
			}
		}
		
		boolean debug = false;
		boolean debug2 = false;
		if(debug2)
		{
			for(int l = 0; l < n; l++)
			{
				for(int r = l; r < n; r++)
				{
					if(pal[l][r] && l != r)
						System.out.println(l + " " + r + " pal");
				}
			}
		}
		
		while(debug)
		{
			int dl = sc.nextInt();
			if(dl == -1)break;
			int dr = sc.nextInt();
			if(dr == -1)break;
			
			System.out.println(dp[dl][dr]);
		}
		
		long res = 0;
		for(int l = 0; l < n; l++)
		{
			for(int r = l; r < n; r++)
			{
				//palin1 is l to r
				
//				if(l > 0)
//					res += dp[0][l-1];
				if(pal[l][r])
				{
					if(r < n-1)
					{
//						System.out.println(l + " " + r + ": " + dp[r+1][n-1]);
						res += dp[r+1][n-1];
					}
				}
			}
		}
		
		System.out.println(res);

	}

}