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.Scanner;
import java.io.OutputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author codeKNIGHT
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		Scanner in = new Scanner(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskD solver = new TaskD();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskD {
	public void solve(int testNumber, Scanner in, PrintWriter out) {
        String s=in.next();
        int i,j,count=0,l=s.length();
        String palins[]=new String[(l*(l+1))/2];
        int pos[][]=new int[palins.length][2];
        int start[]=new int[l];
        for(i=0;i<l;i++)
        {
            pos[count][1]=i;
            count++;

            start[i]++;
        }
        String test;

        int len = s.length(),k;
  for(i=1;i<len-1;i++){
   for(j=i-1,k=i+1;j>=0&&k<len;j--,k++){
    if(s.charAt(j) == s.charAt(k)){
     start[j]++;
        pos[count][1]=k;
        count++;

    }else{
     break;
    }
   }
  }
  //EVEN Occuring Palindromes
  for(i=0;i<len-1;i++){
   for(j=i,k=i+1;j>=0&&k<len;j--,k++){
    if(s.charAt(j) == s.charAt(k)){
     start[j]++;
        pos[count][1]=k;
        count++;
    }else{
     break;
    }
   }
  }
        long c=0;

        for(i=start.length-2;i>=0;i--)
            start[i]+=start[i+1];

        for(i=0;i<count;i++)
            if(pos[i][1]<l-1)
            c+=start[pos[i][1]+1];
        out.println(c);
	}
}