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 D159 {
	
	static public void main(String args[]) {
		Scanner reader = new Scanner(System.in);
		
		String stri = reader.next();
		char[] str = stri.toCharArray();
		int[] beginHere = new int[stri.length()];
		int[] cumulative = new int[stri.length()+2];
		long res = 0;
		for(int i=str.length-1;i>=0;i--) {
			//odd length palindrome
			for(int k=0;i-k >= 0 && i+k <str.length;k++) {
				if(str[i-k] != str[i+k])
					break;
				beginHere[i-k]++;
				res += cumulative[i+k+1];
			}
			for(int k=0;i-k >= 0 && i+k+1 <str.length;k++) {
				if(str[i-k] != str[i+k+1])
					break;
				beginHere[i-k]++;
				res += cumulative[i+k+2];
			}
			cumulative[i] = cumulative[i+1] + beginHere[i];
		}
		System.out.println(res);
	}
}