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.lang.*;
import java.math.BigInteger;


public class Main {
  


	public static void main(String[] args) throws Exception
	{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int t,ca,i,j;
		String[] my;
		String line = br.readLine();
        int n = line.length();
        int[][] ch = new int[n][n];
        for(i=0;i<n;i++)
        ch[i][i] = 1;
        for(i=0;i<n-1;i++)
        {
            if(line.charAt(i)==line.charAt(i+1))ch[i][i+1] = 1;
        }
        int l;
         for(l=3;l  <= n;l++)
        for(i=0;i + l <=n;i++)
        {
          
            
               j = i+l-1;
               if(line.charAt(i) == line.charAt(j))ch[i][j] = ch[i+1][j-1];
            
        }
      //  for(i=0;i<n;i++)
      //  System.out.println(Arrays.toString(ch[i]));
        
        long[] till = new long[n+2];
        long[] end = new long[n+2];
        long ans = 0;
        for(i=1;i<=n;i++)
        {
            long cur = 0;
            for(j=1;j<=i;j++)
            {
                if(ch[j-1][i-1] ==1)cur++;
            }
           // System.out.println("ending at " + j + " total " + cur);
            till[i] = till[i-1] + cur;
        }
        for(i=n;i>=1;i--)
        {
            long cur = 0;
            for(j=i;j<=n;j++)
            if(ch[i-1][j-1] ==1)cur++;
            end[i] = cur;
        }
        for(i=1;i<=n;i++){
   // System.out.println("till " + i +  " total " + till[i] + " and starting at " + (i+1) + " total " + end[i+1]);
      ans+=till[i] * end[i+1];
        }
      System.out.println(ans);
        }
    
}
class ut implements Comparator<Team>{
    public int compare(Team first, Team other)
    {
          if(first.pr > other.pr)return -1;
        if(first.pr < other.pr)return 1;
        if(first.ti < other.ti)return -1;
        if(first.ti > other.ti)return 1;
        if(first.in < other.in)return -1;
        if(first.in > other.in)return 1;
        return 0;
    }
}
class Team implements Comparable{
    int pr,ti,in;
    public Team(int a, int b, int c)
    {
        pr = a;
        ti = b;
        in = c;
    }
    public String toString()
    {
        return pr + " " + ti + " " + in+ "\n";
    }
    public int compareTo(Object obj)
    {
        Team other = (Team)obj;
        if(pr > other.pr)return -1;
        if(pr < other.pr)return 1;
        if(ti < other.ti)return -1;
        if(ti > other.ti)return 1;
        if(in < other.in)return -1;
        if(in > other.in)return 1;
        return 0;
    }
}