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.lang.*;
import java.io.*;
import java.math.*;
 
public class Main
{
	public static void main (String[] args) throws java.lang.Exception
	{
	    Scanner sc = new Scanner(System.in);
		String s = sc.next();
        int n = s.length();
        int firstsum[] = new int[n];
        int[][] A = new int[n][n];
        
        for(int i=0;i<n;++i)
        {
             A[i][i]=1;
        }
        
        for(int len=1;len<n;++len)
        {
            for(int i=0;i<n-len;++i)
            {
                if( s.charAt(i)==s.charAt(i+len) && (i+1>=i+len-1 || A[i+1][i+len-1]==1) )
                {
                    A[i][len+i]=1;
                }
                else{
                    A[i][len+i]=0;
                }
            }
        }
        
        
        for(int i=0;i<n;++i)
        {
            firstsum[i]=0;
            for(int j=0;j<=i;++j)
            {
                if(A[j][i]==1)
                    firstsum[i]++;
            }
            if(i>0)
                firstsum[i]+=firstsum[i-1];
        }
        long cnt=0;
        for(int i=1;i<n;++i)
        {
            for(int j=i;j<n;++j)
                cnt+=A[i][j]*firstsum[i-1];
        }
        System.out.print(cnt);
		sc.close();
	}
	public static int check(String str)
	{
	    int n = str.length();
	    for(int i=0;i<n/2;++i)
	    {
	        if(str.charAt(i)!=str.charAt(n-1-i))
	            return 0;
	    }
	    return 1;
	}
}