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 final class Ideone {
    public static void main(String[] args) {
        int start, end;
        long beforei = 0, totalPair = 0;
        Scanner scanner = new Scanner(System.in);
        char[] mass = scanner.nextLine().toCharArray();
        long[] before = new long[mass.length];
        for (int i = 0 ; i < mass.length; i++) {
            int localtotal = 0;
            out: for (int j = i; j < mass.length; j++) {
                start = i; end = j;
                while (end > start) {
                    if (mass[start++] != mass[end--])
                        continue out;
                }
                localtotal++; before[j]++;
            }
            totalPair+=localtotal*beforei;
            beforei += before[i]; 
       
        }
        System.out.println(totalPair);
    
    }
}