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;

public class Task159D {

    static String s;

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        s = scanner.next();
        int n = s.length();
        long[] r = new long[n];
        long[] l = new long[n];
        
        for (int i = 0; i < n; i++) {
            for (int k = 0; (i - k >= 0) && (i + k < n)
                    && (s.charAt(i - k) == s.charAt(i + k)); k++) {
                r[i + k]++;
                l[i - k]++;
            }
            for (int k = 0; (i - k >= 0) && (i + k + 1 < n)
                    && (s.charAt(i - k) == s.charAt(i + k + 1)); k++) {
                r[i + k + 1]++;
                l[i - k]++;
            }
        }

        long sum = 0;
        for (int i = 0; i < n - 1; i++) {
            for (int j = i + 1; j < n; j++) {
                sum += r[i] * l[j];
            }
        }

        System.out.println(sum);
    }

}