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.math.*;
import java.io.*;
public class Main
	{
	public static void main(String args[])	throws IOException
		{
		BufferedReader c=new BufferedReader(new InputStreamReader(System.in));
		//Scanner c=new Scanner(System.in);
		char S[]=c.readLine().toCharArray();
		int N=S.length;
		boolean dp[][]=new boolean[N][N];
		for(int i=N-1;i>=0;i--) 
			{
			dp[i][i]=true;
			if(i<N-1)
				{
				if(S[i]==S[i+1])
					dp[i][i+1]=true;
				else
					dp[i][i+1]=false;
				}
			for(int j=i+2;j<N;j++) 
				{
				if(S[i]==S[j]&&dp[i+1][j-1])
					dp[i][j]=true;
				}
			}
		int ending_at[]=new int[N];			//number of palindromes ending strictly before i
		for(int i=0;i<N;i++) 
			{
			for(int j=i;j<N-1;j++)		//end has to be less than N-1 
				{
				if(dp[i][j])			//i to j is a palindrome
					ending_at[j]++;
				}
			}
		int ending_bef[]=new int[N];			//number of palindromes ending strictly before i
		for(int i=1;i<N;i++) 
			{
			ending_bef[i]=ending_bef[i-1]+ending_at[i-1];
			}
		long ans=0;
		for(int i=1;i<N;i++)				//2nd palindrome 
			{
			for(int j=i;j<N;j++) 
				{
				if(dp[i][j])
					ans+=ending_bef[i];
				}
			}
		System.out.println(ans);
		}
	}


//must declare new classes here