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.*;

public class palin_pairs 
{
	static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }
       
	}
	public String revString (String s)
	{
		String st = "";
		for (int i = 0; i< s.length(); i++)
		{
			st = s.charAt(i)+st;
		}
		return st;
	}
	public boolean checkPalin(String s)
	{
		String str = revString(s);
		if (str.equals(s))
			return true;
		
		else return false;
			
	}
	
	public static void main(String args[])throws IOException
	{
		InputStream inputstream = System.in;
		OutputStream outputstream = System.out;
		InputReader sc = new InputReader(inputstream);
		PrintWriter out = new PrintWriter(outputstream);
		palin_pairs pp = new palin_pairs();
		
		
		String s = sc.next();
		int n = s.length();
		int [][] dp = new int [n][n];
    // dp solution for finding palindrome solution  //
		
		for (int i = 0; i< n; i++)
			dp[i][i]=1;
		
		for (int i = 0; i<n-1; i++)
		{
			if (s.charAt(i)==s.charAt(i+1))
				dp[i][i+1]=1;
		}
		for (int len = 3; len<=n; len++)
		{
			for (int i = 0; i< n-len+1; i++)
			{
				int j = i+len-1;
				if ((s.charAt(i)==s.charAt(j))&&(dp[i+1][j-1]==1))
						dp[i][j]=1;
			}
		}
		
		
		/// cumulative sums /////////
		int [][] dprow= new int [n][n]; 
		for (int i = 0; i< n; i++)
		{
			dprow[i][0] = dp[i][0];
			for (int j = 1; j< n; j++)
			{
				dprow[i][j]=dprow[i][j-1]+dp[i][j];
				
				
			}
		}
		for (int j = 0;j < n; j++)
		{
			for (int i = n-1; i >= 1; i--)
			{
				dprow[i-1][j] += dprow[i][j];
			}
		}
		//// cumulative sums //////////
		
		long sum = 0;
		for (int i = 0; i< n; i++)
		{
			for (int j = 0; j< n; j++)
			{
				if (dp[i][j]==1)
				{
					if (i-1>=0)
						sum = sum + dprow[0][i-1];
					if (j+1<n)
						sum = sum + dprow[j+1][n-1];
					
				}
					
			}
		}
		out.println(sum>>>1);
		out.close();
		out.flush();
	}

}