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.PrintWriter;
import java.util.Scanner;

public class Main {

   Scanner in;
   PrintWriter out;

   void asserT(boolean e) {
      if (!e) {
         throw new Error();
      }
   }

   long startFrom[];
   long endAt[];
   char ch[];

   void calcPal(int left,int right){
      while(true){
         if (left>=0 && right<ch.length && ch[left]==ch[right]){
            startFrom[left]++;
            endAt[right]++;
            left--;
            right++;
         }else{
            break;
         }
      }
   }

   void solve() {
      ch = in.nextLine().toCharArray();
      startFrom = new long[ch.length];
      endAt = new long[ch.length];
      for (int i = 0; i < ch.length; i++) {
         calcPal(i, i);
         if (i + 1 < ch.length) {
            calcPal(i, i + 1);
         }
      }
      long ans = 0;
      for (int end = 0; end < ch.length; end++) {
         for (int start = end + 1; start < ch.length; start++) {
            ans += endAt[end] * startFrom[start];
         }
      }
      out.println(ans);
   }

   void run() {
      in = new Scanner(System.in);
      out = new PrintWriter(System.out);
      try {
         solve();
      } finally {
         out.close();
      }
   }

   public static void main(String[] args) {
      new Main().run();
   }
}