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 D {
    public static void main(String args[]) {
        Scanner in = new Scanner(System.in);
        String s = in.nextLine();
        int odd[] = new int[s.length()];
        int even[] = new int[s.length()];
        int l = 0;
        int r = -1;
        for (int i = 0; i < s.length(); i++) {
            int k;
            if (i > r) {
                k = 0;
            } else {
                k = Math.min(odd[l + r - i], r - i);
            }
            k++;
            while (i + k < s.length() && i - k >= 0 && s.charAt(i + k) == s.charAt(i - k)) {
                k++;
            }
            k--;
            odd[i] = k;
            if (i + k > r) {
                l = i - k;
                r = i + k;
            }
        }
        l = 0;
        r = -1;
        for (int i = 0; i < s.length(); i++) {
            int k;
            if (i > r) {
                k = 0;
            } else {
                k = Math.min(even[l + r - i + 1], r - i + 1);
            }
            k++;
            while ((i + k - 1) < s.length() && i >= k && s.charAt(i + k - 1) == s.charAt(i - k)) {
                k++;
            }
            k--;
            even[i] = k;
            if (i + k - 1 > r) {
                l = i - k;
                r = i + k - 1;
            }
        }
        int start[] = new int[s.length()];
        int finish[] = new int[s.length()];
        for (int i=0;i<s.length();i++) {
            for (int j=0;j<=odd[i];j++)
            {
                start[i-j]++;
                finish[i+j]++;
            }
            for (int j=1;j<=even[i];j++) {
                start[i-j]++;
                finish[i+j-1]++;
            }
        }
        long ans = 0;
        for (int i=0;i<s.length();i++)
        {
            long count = 0;
            for (int j=(i+1);j<s.length();j++)
                count+=start[j];
            ans+=(count * (long)finish[i]);
        }
        System.out.println(ans);
    }
}