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.io.*;
import java.util.*;
import java.math.*;

public class Main
{
	BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	StringTokenizer tokenizer=null;
	
	public static void main(String[] args) throws IOException
	{
		new Main().execute();
	}
	
	void debug(Object...os)
	{
		System.out.println(Arrays.deepToString(os));
	}
	
	int ni() throws IOException
	{
		return Integer.parseInt(ns());
	}
	
	long nl() throws IOException 
	{
		return Long.parseLong(ns());
	}
	
	double nd() throws IOException 
	{
		return Double.parseDouble(ns());
	}
		
	String ns() throws IOException 
	{
		while (tokenizer == null || !tokenizer.hasMoreTokens()) 
			tokenizer = new StringTokenizer(br.readLine());
		return tokenizer.nextToken();
	}
	
	String nline() throws IOException
	{
		tokenizer=null;
		return br.readLine();
	}
		
	
	//Main Code starts Here
	int totalCases, testNum;	
	String s;
	int[][] dp;
	void execute() throws IOException
	{
		totalCases = 1;
		for(testNum = 1; testNum <= totalCases; testNum++)
		{
			input();
			solve();
		}
	}

	void solve()
	{
		int n = s.length();
		dp = new int[n+2][n+2];
		
		for(int i=1;i<=n;i++)
			for(int j = 1;j<=n;j++)
				if(i==j)
					dp[i][i]=1;
				else 
					dp[i][j]=0;
					
		for(int i = 1;i<n;i++)
		{
			if(s.charAt(i-1)==s.charAt(i))
				dp[i][i+1]=1;
		}
				
		for(int j = 1;j<=n;j++)
			for(int i = 1;i<=n;i++)
				if(i<j && dp[i+1][j-1]==1 && s.charAt(i-1)==s.charAt(j-1))
					dp[i][j] = 1;
					
		int[] colcount = new int[n+2];
		for(int j=1;j<=n;j++)
			for(int i = 1;i<=n;i++)
				colcount[j]+=dp[i][j];
		
		
		//debug(colcount);
		
		long[] rowcount = new long[n+2];
		for(int j=1;j<=n;j++)
			for(int i = 1;i<=n;i++)
				rowcount[j]+=dp[j][i];
		
		for(int i = n-1;i>=1;i--)
			rowcount[i]+=rowcount[i+1];
					
		//debug(rowcount);
					
		long sum=0;
		for(int i =1;i<n;i++)
		{
			sum+=rowcount[i+1]*colcount[i];
			//debug(sum);
		}
			
		System.out.println(sum);
		
		/*for(int i = 1;i<=n;i++)
		{
			for(int j=1;j<=n;j++)
					System.out.print(dp[i][j]+" ");
			System.out.println();
		}*/
			
	}
	

	void printarr(int [] a,int b)
	{
		for(int i = 0;i<=b;i++)
		{
			if(i==0)
				System.out.print(a[i]);
			else
				System.out.print(" "+a[i]);
		}
		System.out.println();
	}
	
	boolean input() throws IOException
	{
		s = ns();
		return true;
	}
}