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.*;

public class PalindromePairs {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		String word = sc.next();
		
		ArrayList<Palindrome> pals = new ArrayList<Palindrome>();
		
		// Assemble list of all palindrome start/end points.		
		for (int i = 0; i < word.length(); i++)
		{
			int leftBound = i, rightBound = i;
			while ((leftBound >= 0) && (rightBound < word.length()))
			{
				if (word.charAt(leftBound) == word.charAt(rightBound))
				{
					pals.add(new Palindrome(leftBound,rightBound));
				}
				else
				{
					break;
				}
				leftBound -= 1;
				rightBound += 1;
			}
			
			if (i+1 != word.length())
			{
				leftBound = i;
				rightBound = i+1;
				
				while ((leftBound >= 0) && (rightBound < word.length()))
				{
					if (word.charAt(leftBound) == word.charAt(rightBound))
					{
						pals.add(new Palindrome(leftBound,rightBound));
					}
					else
					{
						break;
					}
					leftBound -= 1;
					rightBound += 1;
				}
			}
		}
		
		Collections.sort(pals);
		Collections.reverse(pals);
		
		int[] num_ended = new int[pals.size()]; // Array to track number of palindromes that have ended at this point.
		int pal_index = 0;
		for (int i = 0; i < pals.size(); i++)
		{
			if (i != 0)
			{
				num_ended[i] = num_ended[i-1];
			}
			
			boolean done = false;
			while (!done)
			{
				if (pal_index >= pals.size())
				{
					break;
				}
				
				if (pals.get(pal_index).end == i-1)
				{
					num_ended[i] += 1;
					pal_index += 1;
				}
				else
				{
					done = true;
				}
			}
		}
		
		long pal_count = 0;
		for (Palindrome p: pals)
		{
			pal_count += num_ended[p.start];
		}
		
		System.out.println(pal_count);
	}
	
	private static class Palindrome implements Comparable<Palindrome>
	{
		int start;
		int end;
		
		public Palindrome(int start, int end)
		{
			this.start = start;
			this.end = end;
		}

		@Override
		public int compareTo(Palindrome o) {
			// TODO Auto-generated method stub
			return o.end - this.end;
		}
	}
}