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.StringTokenizer;

public class D {
	BufferedReader in;
	StringTokenizer str;
	PrintWriter out;
	String SK;

	String next() throws IOException {
		while ((str == null) || (!str.hasMoreTokens())) {
			SK = in.readLine();
			if (SK == null)
				return null;
			str = new StringTokenizer(SK);
		}
		return str.nextToken();
	}

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

	double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}

	long nextLong() throws IOException {
		return Long.parseLong(next());
	}

	void solve() throws IOException {
		char[] a = next().toCharArray();
		int n = a.length;
		boolean ispal[][] = new boolean[n][n];
		for (int i = 0; i < n; i++) {
			for (int j = 0; j <= i; j++)
				ispal[i][j] = true;
		}
		int j;
		for (int len = 2; len < n; len++) {
			for (int i = 0; i + len - 1 < n; i++) {
				j = i + len - 1;
				ispal[i][j] = (a[i] == a[j]) && ispal[i + 1][j - 1];
			}
		}
		int res[] = new int[n];
		long resul = 0;
		for (int i = 0; i < n; i++) {
			for (j = 1; j <= i; j++) {
				if (ispal[j][i]) {
					resul = resul + (long) res[j - 1];
				}
			}
			if (i > 0) {
				res[i] = res[i - 1];
			}
			for (j = 0; j <= i; j++) {
				if (ispal[j][i]) {
					res[i]++;
				}
			}
		}
		out.println(resul);
	}

	void run() throws IOException {
		in = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		solve();
		out.close();
	}

	public static void main(String[] args) throws IOException {
		new D().run();
	}

}