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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

/**
 *
 * @author evgeniy
 */
public class DTask {

    public static int min(int a, int b) {
        return (a < b ? a : b);
    }

    public static void main(String[] args) {
        try {
            BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
            String s = br.readLine();
            while (s.charAt(s.length() - 1) > 'z' || s.charAt(s.length() - 1) < 'a') {
                s = s.substring(0, s.length() - 1);
            }

            int l = 0;
            int r = -1;
            int n = s.length();
            int[] d1 = new int[n];
            int[] d2 = new int[n];
            for (int i = 0; i < n; i++) {
                int k = (i > r ? 0 : min(d1[l + r - i], r - i)) + 1;
                while (i + k < n && i - k >= 0 && s.charAt(i + k) == s.charAt(i - k)) {
                    ++k;
                }
                d1[i] = --k;
                if (i + k > r) {
                    l = i - k;
                    r = i + k;
                }
            }
            l = 0;
            r = -1;
            for (int i = 0; i < n; i++) {
                d1[i]++;

                int k = (i > r ? 0 : min(d2[l + r - i + 1], r - i + 1)) + 1;
                while (i + k - 1 < n && i - k >= 0 && s.charAt(i + k - 1) == s.charAt(i - k)) {
                    ++k;
                }
                d2[i] = --k;
                if (i + k - 1 > r) {
                    l = i - k;
                    r = i + k - 1;
                }
            }

            int[] num = new int[n];
            for (int i = 0; i < n; i++) {
                int alf = 0;
                for (int k = i; k > i - d1[i]; k--) {
                    alf++;
                    num[k] += alf;
                }

                for (int k = i - d1[i]; k >= 0; k--) {
                    num[k] += alf;
                }
                alf = 0;
                for (int k = i - 1; k >= i - d2[i]; k--) {
                    alf++;
                    num[k] += alf;
                }
                for (int k = i - d2[i] - 1; k >= 0; k--) {
                    num[k] += alf;
                }
            }

            long counter = 0;
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < d1[i]; j++) {
                    if (i + j + 1 < n) {
                        counter += num[i + j + 1];
                    }
                    /*for (int k = i + j + 1; k < n; k++) {
                    counter += (long) min(d1[k], k - i - j);
                    }
                    for (int k = i + j + 2; k < n; k++) {
                    counter += (long) min(d2[k], k - i - j - 1);
                    
                    }*/
                }
                for (int j = 0; j < d2[i]; j++) {
                    if (i + j + 1 < n) {
                        counter += num[i + j + 1];/*for (int k = i + j + 1; k < n; k++) {
                        counter += (long) min(d1[k], k - i - j);
                        }
                        for (int k = i + j + 2; k < n; k++) {
                        counter += (long) min(d2[k], k - i - j - 1);
                        
                        }*/
                    }
                }
            }
            System.out.println(counter);
        } catch (IOException ex) {
        }
    }
}