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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Qualification2_D {
	public static void main(String[] args) throws IOException {
		BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
		String s = r.readLine();
		char[] arr = s.toCharArray();
		boolean[][] isPalindrome = new boolean[arr.length][arr.length];
		for (int i = 0; i < arr.length; i++) {
			isPalindrome[i][i] = true;
			int start = i - 1, end = i + 1;
			while (start >= 0 && end < arr.length && arr[start] == arr[end]) {
				isPalindrome[start][end] = true;
				start--;
				end++;
			}
		}
		for (int i = 0; i < arr.length; i++) {
			if (i + 1 < arr.length && arr[i] == arr[i + 1]) {
				isPalindrome[i][i + 1] = true;
				int start = i - 1, end = i + 2;
				while (start >= 0 && end < arr.length && arr[start] == arr[end]) {
					isPalindrome[start][end] = true;
					start--;
					end++;
				}
			}
		}

		long[] count = new long[arr.length];
		for (int i = 0; i < count.length; i++) {
			for (int j = i; j < count.length; j++) {
				if (isPalindrome[i][j])
					count[i]++;
			}
		}
		for (int i = count.length - 2; i >= 0; i--) {
			count[i] += count[i + 1];
		}
		long res = 0;
		for (int i = 0; i < isPalindrome.length; i++) {
			for (int j = i; j < isPalindrome.length; j++) {
				if (isPalindrome[i][j]) {
					if (j + 1 < arr.length)
						res += count[j + 1];
				}
			}
		}
		System.out.println(res);
	}
}