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.*;

public class PalindromePairs{
   public static void main(String[] Args){
      Scanner sc = new Scanner(System.in);
      String s = sc.next();
      long[] counts = new long[s.length()]; 
      long[] rc = new long[s.length()]; 
      int n = s.length();
      boolean[][] ispal = new boolean[n][n];
      for(int k =0;k<n;k++){
         ispal[k][k] = true;
         if(k!=0)
            ispal[k][k-1] = true;
      }
      for(int k = 1;k<n;k++){
         for(int j = 0; j+k < n; j++){
            if(s.charAt(j)==s.charAt(j+k))
               ispal[j][j+k] = ispal[j+1][j-1+k];
         }
      }
      for(int k = 0;k<n;k++)
         for(int j = 0;j<=k;j++)
            if(ispal[j][k]){
               counts[k]++;
               rc[j]++;
            }
      long ans = 0;
      for(int k = 1;k<n;k++){
         counts[k] += counts[k-1];
      }
      for(int k =0;k<n-1;k++)
         ans+= counts[k] * rc[k+1];
      System.out.println(ans);
   }
}