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.File;
import java.io.FileNotFoundException;
import java.util.Scanner;

/**
 * @author konovalov_nik
 */
public class D {
    public static void main(String[] args) throws FileNotFoundException {
        Scanner scanner = new Scanner(System.in);
        //Scanner scanner = new Scanner(new File("input.txt"));
        
        String s = scanner.next();
        
        long[] l = new long[s.length()];
        long[] r = new long[s.length()];
        int len = s.length();
                
        for (int i = 0; i < len; i++) {
            int k = 0;
            while (i - k >= 0 && i + k < len) {
                if (s.charAt(i - k) == s.charAt(i + k)) {
                    l[i + k]++;
                    r[i - k]++;
                    k++;
                } else {
                    break;
                }
            }
        }

        for (int i = 0; i < len; i++) {
            int k = 0;
            while (i - k >= 0 && i + k + 1 < len) {
                if (s.charAt(i - k) == s.charAt(i + k + 1)) {
                    l[i + k + 1]++;
                    r[i - k]++;
                    k++;
                } else {
                    break;
                }
            }
        }
        
        long[] suml = new long[l.length];
        long[] sumr = new long[r.length];
        
        suml[0] = l[0];
        for (int i = 1; i < len; i++) {
            suml[i] = suml[i - 1] + l[i];
        }

        sumr[len - 1] = r[len  -1];

        for (int i = len - 2; i >= 0; i--) {
            sumr[i] = sumr[i + 1] + r[i];
        }
        
        long ans = 0;

        for (int i = 0; i < len; i++) {
            int k = 0;
            while (i - k >= 0 && i + k < len) {
                if (s.charAt(i - k) == s.charAt(i + k)) {
                    if (i - k - 1 >= 0) {
                        ans += suml[i - k -1];
                    }
                    if (i + k + 1 < len) {
                        ans+= sumr[i + k + 1];
                    }
                    k++;
                } else {
                    break;
                }
            }
        }

        for (int i = 0; i < len; i++) {
            int k = 0;
            while (i - k >= 0 && i + k + 1 < len) {
                if (s.charAt(i - k) == s.charAt(i + k + 1)) {
                    if (i - k - 1 >= 0) {
                        ans += suml[i - k -1];
                    }
                    if (i + k + 2 < len) {
                        ans+= sumr[i + k + 2];
                    }
                    k++;
                } else {
                    break;
                }
            }
        }

        System.out.println(ans / 2);
        
    }
    
}