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 MainClass {

	public static int palinrome(String str) {
		int n = str.length();
		int[] d1 = new int[str.length()];
		int l = 0, r = -1;
		
		for (int i = 0; i < n; ++i) {
			int k = (i > r ? 0 : Math.min(d1[l + r - i], r - i)) + 1;
			while (i + k < n && i - k >= 0 && str.charAt(i+k) == str.charAt(i-k)) {
				++k;
			}
			d1[i] = --k;
			if (i + k > r) {
				l = i - k;
				r = i + k;
			}
		}
	
		int[] d2 = new int[str.length()];
		
		l = 0;
		r = -1;
		for (int i = 0; i < n; ++i) {
			int k = (i > r ? 0 : Math.min(d2[l + r - i + 1], r - i + 1)) + 1;
			while (i + k - 1 < n && i - k >= 0 && str.charAt(i+k-1) == str.charAt(i - k)) {
				++k;
			}
			d2[i] = --k;
			if (i + k -1 > r) {
				l = i - k; 
				r = i + k - 1;
			}
		}
		
		int answer = 0;
		
		for (int i = 0; i < n; i++) {
			answer += d1[i] + d2[i];
		}
		
		return answer;
	}
	
	private String str = new String();
	private long answ = 0;
	
	public static void main(String[] args) {
		new MainClass().run();
	}
	
	public void run() {
		Scanner in = new Scanner(System.in);
		str = in.next();
		int n = str.length();
		int prevNum1 = 0, prevNum2 = 0;
		for (int i = 0; i < n - 1; i++) {
			int p1 = palinrome(str.substring(0, i + 1));
			int p2 = palinrome(str.substring(i + 1, n));
			answ += p2;
			answ += (p1 - prevNum1) * p2 + p1 * (Math.max(p2 - prevNum2, 0)) + (p1 - prevNum1) * str.substring(i + 1, n).length();
			prevNum1 = p1;
			prevNum2 = p2;
			answ += i + 1;
		}
		System.out.println(answ);
	}
}