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.util.*;
import java.io.*;

public class D159 
{
	public static void main(String[] args) { new D159(); }
	FS in = new FS();
	PrintWriter out = new PrintWriter(System.out);
	
	int n;
	char[] str;
	boolean[][] isPal;
	int[] pref, suff;
	
	D159() {
		n = (str = in.next().toCharArray()).length;
		isPal = new boolean[n][n];
		
		pref = new int[n];
		for (int i = 0; i < n; i++)
			for (int j = i; j >=0; j--) {
				if (str[j] == str[i] && ((i - j < 2) || isPal[j + 1][i - 1]))
					isPal[j][i] = true;
				if (isPal[j][i])
					pref[i]++;
			}
		
		suff = new int[n];
		for (int i = n - 1; i >= 0; i--) {
			if (i < n - 1)
				suff[i] = suff[i + 1];
			for (int j = i; j < n; j++)
				if (isPal[i][j])
					suff[i]++;
		}

		long ans = 0;
		for (int i = 0; i < n - 1; i++)
			ans += pref[i] * suff[i + 1];
		out.println(ans);

		out.close();
	}
	
	

	int min(int a, int b) { if (a < b) return a; return b; }	
	int max(int a, int b) { if (a > b) return a; return b; }	
	long min(long a, long b) { if (a < b) return a; return b; }	
	long max(long a, long b) { if (a > b) return a; return b; }	

	boolean z(int x) { if (x == 0) return true; return false; }
	boolean z(long x) { if (x == 0) return true; return false; }
	
	class FS {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens()) {
				try { st = new StringTokenizer(br.readLine()); }
				catch (Exception e) {}
			} return st.nextToken();
		}
		int nextInt() { return Integer.parseInt(next()); }
		long nextLong() { return Long.parseLong(next()); }
		double nextDouble() { return Double.parseDouble(next()); }
	}
}