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.Scanner;

public class D_159_Palindromic_Pairs {
	public static void main(String[] args){
		Scanner input=new Scanner(System.in);
		String tmp=input.next();
		char[] s=tmp.toCharArray();
		int i,j,k;
		long count=0;
		int[] front=new int[2000],rear=new int[2000];

		for(i=0;i<s.length;i++){
			for(j=i,k=i;j>=0&&k<s.length&&s[k]==s[j];j--,k++){
				front[j]++;rear[k]++;
			}
			for(j=i,k=i+1;j>=0&&k<s.length&&s[k]==s[j];j--,k++){
				front[j]++;rear[k]++;
			}
		}
		for(i=s.length-1;i>0;i--)
			front[i-1]+=front[i];
		for(i=0;i<s.length-1;i++)
			count+=rear[i]*front[i+1];
		System.out.println(count);
	}
}