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;
import java.util.Arrays;

/**
 * User: artpetroff
 * Date: 17.04.12
 * Time: 15:46
 */
public class _159D {

    public static void main(String[] args) {
        try {
            final BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
            final String s = reader.readLine();

            long c = c(s);

            System.out.println(c);

        } catch (IOException e) {
            throw new RuntimeException(e);
        }
    }

    public static long c(final String s) {
        final int n = s.length();

        final int[] d1 = new int[n];
        final int[] d2 = new int[n];
        final long[] l = new long[n];
        final long[] r = new long[n];

        for (int i=0; i<n; ++i)
        {
            d1[i] = 1;
            while (i-d1[i] >= 0 && i+d1[i] < n && s.charAt(i-d1[i]) == s.charAt(i+d1[i]))
                ++d1[i];

            d2[i] = 0;
            while (i-d2[i]-1 >= 0 && i+d2[i] < n && s.charAt(i-d2[i]-1) == s.charAt(i+d2[i]))
                ++d2[i];
        }

        final int[] r1 = Arrays.copyOf(d1, d1.length);
        final int[] r2 = Arrays.copyOf(d2, d2.length);
//        final int[] r1 = Arrays.copyOf(d1, d1.length);
//        final int[] r2 = Arrays.copyOf(d2, d2.length);

        for (int i = 0; i < n; i++) {
            int k = 0;
            while (i + k < n && d1[i] != 0) {
                l[i + k]++;
                d1[i]--;
                k++;
            }
            k = 0;
            while (i + k < n && d2[i] != 0) {
                l[i + k]++;
                d2[i]--;
                k++;
            }
        }
        for (int i = n - 1; i >= 0; i--) {
            int k = 0;
            while (i - k >= 0 && r1[i] != 0) {
                r[i - k]++;
                r1[i]--;
                k++;
            }
            k = 1;
            while (i - k >= 0 && r2[i] != 0) {
                r[i - k]++;
                r2[i]--;
                k++;
            }
        }
        long res = 0;
        for (int i = 0; i < n - 1; i++) {
            for (int j = i + 1; j < n; j++) {
                res += r[j] * l[i];
            }
        }

        return res;
    }

    private static void incD(final String s, final int n, final int[] d, final int i, final int startValue) {
        d[i] = startValue;
        while (i - d[i] >= 0 && i + d[i] < n && s.charAt(i - d[i]) == s.charAt(i + d[i])) {
            d[i]++;
        }
    }

    private static boolean isPalyndrom(final String s, final int i, final int j) {
        for (int k = 0, l = (j - i) / 2; k <= l; k++) {
            if (s.charAt(i + k) != s.charAt(j - k)) {
                return false;
            }
        }
        return true;
    }

}