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 problemD_q2 implements Runnable {
	public StringTokenizer st;
	public BufferedReader in;
	public PrintWriter out;

	public static void main(String[] args) {
		new Thread(new problemD_q2()).start();
	}

	public void run() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(new OutputStreamWriter(System.out));
			solve();
		} catch (Exception e) {
			e.printStackTrace();
			System.exit(9000);
		} finally {
			out.flush();
			out.close();
		}
	}

	public String nextToken() throws IOException {
		while (st == null || !st.hasMoreTokens()) {
			st = new StringTokenizer(in.readLine());
		}
		return st.nextToken();
	}

	public int nextInt() throws NumberFormatException, IOException {
		return Integer.parseInt(nextToken());
	}

	public long nextLong() throws NumberFormatException, IOException {
		return Long.parseLong(nextToken());
	}

	public double nextDouble() throws NumberFormatException, IOException {
		return Double.parseDouble(nextToken());
	}

	public void solve() throws NumberFormatException, IOException {
		String s = nextToken();
		int n = s.length();
		boolean[][] pal = new boolean[n][n+1];
		//System.out.print(pal[0][0]); //== false
		for (int i = 0; i < n; i++) {
			pal[i][0] = true;
			pal[i][1] = true;
		}
		for (int i = n-2; i >= 0; i--) {
			for (int j = 2; i+j-1 < n; j++) {
				if (pal[i+1][j-2] && s.charAt(i) == s.charAt(i+j-1)) pal[i][j] = true;
			}
		}
		int[] suffsum = new int[n+1];
		for (int i = n-1; i > 0; i--) {
			int res = 0;
			for (int j = 1; i+j-1 < n; j++) {
				if (pal[i][j]) res++;
			}
			suffsum[i] = suffsum[i+1] + res;	
		}
		int[] ends = new int[n];
		for (int i = 0; i < n; i++) {
			for (int j = 1; i+j-1 < n; j++) {
				if (pal[i][j]) ends[i+j-1]++;
			}
		}
		long res = 0;
		for (int i = 0; i < n-1; i++) {
			res += ends[i] * suffsum[i+1];
		}
		out.print(res);
	}
}