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.*;
import java.util.*;

public class D159 {
	static String next() {
		if (st == null || !st.hasMoreTokens())
			nextLine();
		return st.nextToken();
	}

	static int nextInt() {
		return Integer.parseInt(next());
	}

	static void nextLine() {
		try {
			st = new StringTokenizer(f.readLine());
		} catch (Exception e) {
			e.printStackTrace();
		}
	}

	static StringTokenizer st;
	static PrintWriter out;
	static BufferedReader f;

	public static void main(String[] args) throws IOException {
		f = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		char[] s = next().toCharArray();
		int N = s.length;
		long[] start = new long[N], end = new long[N];
		for (int mid = 0; mid < N; mid++) {
			for (int off = 0; mid - off >= 0 && mid + off < N
					&& s[mid - off] == s[mid + off]; off++) {
				start[mid - off]++;
				end[mid + off]++;
			}
		}
		for (int mid = 0; mid + 1 < N; mid++) {
			int mid1 = mid + 1;
			for (int off = 0; mid - off >= 0 && mid1 + off < N
					&& s[mid - off] == s[mid1 + off]; off++) {
				start[mid - off]++;
				end[mid1 + off]++;
			}
		}

		long sum = 0;
		for (int x = 0; x < N; x++) {
			for (int y = x + 1; y < N; y++) {
				sum += end[x] * start[y];
			}
		}
		out.println(sum);
		out.close();
		// System.exit(0);
	}

}