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.InputStreamReader;

public class D159_2 {
    static long[] d1Total;
    static long[] d2Total;
    public static void main(String[] args) throws Exception {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        String s = in.readLine();
        d1Total = countD1(s);
        d2Total = countD2(s);
        long res = 0;
        int n = s.length();
        long[] am = new long[n];
        for (int i = 0 ; i < n ; i++) {
            am[i] = 1;
            for (int j = 1 ; j <= d1Total[i] ; j++)
                am[i - j]++;
            for (int j = 0 ; j < d2Total[i] ; j++)
                am[i - j - 1]++;
        }
        long[] t = new long[n];
        t[n - 1] = am[n - 1];
        for (int i = n - 2 ; i >= 0 ; i--) {
            t[i] = t[i + 1] + am[i];
        }
        for (int i = 0 ; i < n ; i++) {
            for (int j = 0 ; j <= d1Total[i] ; j++)
                if (i + j + 1 < n) res += t[i + j + 1];
            for (int j = 0 ; j < d2Total[i] ; j++)
                if (i + j + 1 < n) res += t[i + j + 1];
        }
        System.out.println(res);
    }

    public static long[] countD1(String s) {
        int n = s.length();
        long[] res = new long[n];
        int l = 0;
        int r = -1;
        for (int i = 0; i < n; i++) {
            int k = (int) ((i > r ? 0 : Math.min(res[l + r - i], r - i)) + 1);
            while (i + k < n && i - k >= 0 && s.charAt(i + k) == s.charAt(i - k)) {
                ++k;
            }
            res[i] = --k;
            if (i + k > r) {
                l = i - k;
                r = i + k;
            }
        }
        return res;
    }

    public static long[] countD2(String s) {
        int n = s.length();
        long[] res = new long[n];
        int l = 0;
        int r = -1;
        for (int i = 0; i < n; i++) {
            int k = (int) ((i > r ? 0 : Math.min(res[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;
            }
            res[i] = --k;
            if (i + k - 1 > r) {
                l = i - k;
                r = i + k - 1;
            }
        }
        return res;
    }
}