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.io.PrintWriter;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		PrintWriter out = new PrintWriter(System.out, true);
		Scanner in = new Scanner(System.in);
		Solver solver = new Solver(in, out);
		solver.run();
	}
}

class Solver {
	Scanner     in;
	PrintWriter out;

	public Solver(Scanner i, PrintWriter o) {
		in  = i;
		out = o;
	}
	
	int[] left  = new int[2000];
	int[] right = new int[2000];

	public void run() {
		String s = in.next();
		int n = s.length();
		int i, j, k;
		double ans = 0;
		for (i = 0; i < n; i++) {
			left[i]  = 0;
			right[i] = 0;
		}
		
		for (i = 0; i < n; i++) {
			j = i;
			k = i;
			while (j >= 0 && k < n) {
				if (s.charAt(j) != s.charAt(k)) break;
				left[k]++;
				right[j]++;
				j--;
				k++;
			}
			j = i-1;
			k = i;
			while (j >= 0 && k < n) {
				if (s.charAt(j) != s.charAt(k)) break;
				left[k]++;
				right[j]++;
				j--;
				k++;
			}
		}
		for (i = 0; i < n-1; i++) {
			for (j = i+1; j < n; j++) {
				ans += (left[i]*right[j]);
			}
		}		
		out.format("%.0f", ans);
	}	
}