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:
/* package codechef; // don't place package name! */

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

/* Name of the class has to be "Main" only if the class is public. */
public class Codechef
{
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		Scanner sc = new Scanner(System.in);
		String str = sc.next();
		int n = str.length();
		boolean dp[][] = new boolean[n][n];
		int end[] = new int[n];
		int start[] = new int[n];
		for(int i=0;i<n;i++){
		    dp[i][i]=true;
		    end[i]=1;
		    start[i]=1;
		}
		for(int l=2;l<n+1;l++){
		    for(int i=0;i<n-l+1;i++){
		        int j = i+l-1;
		        if(l==2){
		            if(str.charAt(i)==str.charAt(j)){
		                dp[i][j]=true;
		                start[i]++;
		                end[j]++;
		            }
		        }
		         else{
		                if(str.charAt(i)==str.charAt(j)&&dp[i+1][j-1]){
		                    dp[i][j]=true;
		                    start[i]++;
		                    end[j]++;
		                }
		            }
		    }
		}
		for(int i =(n-2);i>=0;i--){
		    start[i]+=start[i+1];
		}
		long res=0;
		
// 		for(int i=0;i<n;i++){
// 		    System.out.println(start[i]+" "+end[i]);
// 		}
		for(int i=0;i<n-1;i++){
		    res += end[i]*start[i+1];
		}
		System.out.println(""+res);
		
	}
}