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 d159 {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		char[] word = in.next().toCharArray();
		long total = 0;
		long paliBefore = 0;
		long[] paliEndedBy = new long[word.length];
		for (int i = 0; i < word.length; i++) {
			long pali = 0;
			outer: for (int j = i; j < word.length; j++) {
				int a = i, b = j;
				while (b > a) {
					if (word[a++] != word[b--])
						continue outer;
				}
				pali++;
				paliEndedBy[j]++;
			}
			total += pali * paliBefore;
			paliBefore += paliEndedBy[i];
		}
		
		System.out.println(total);
	}
}