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.*;
public class PalindromePairs{
	public static void main(String [] args){
		final Scanner s = new Scanner(System.in);
		final char [] c = s.nextLine().toCharArray();
		final int l = c.length;
		final int [][] dp1 = new int[l][l];
		for (int i = 0; i < l; ++i){
			for (int j = 0; j + i < l; ++j){
				if (c[j] == c[j + i]){
					dp1[j][j + i] = 1;
					if (j + 1 < j + i - 1){
						dp1[j][j + i] = dp1[j + 1][j + i - 1];
					}
				}
			}
		}
		final int [] dp2 = new int[l];
		final int [] dp3 = new int[l];
		for (int i = 0; i < l; ++i){
			for (int j = 0; j <= i; ++j){
				dp2[i] += dp1[j][i];
				dp3[j] += dp1[j][i];
			}
		}
		long res = 0;
		for (int i = 0; i < l; ++i){
			for (int j = i + 1; j < l; ++j){
				res += dp2[i]*dp3[j];
			}
		}
		System.out.println(res);
	}
}