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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskD solver = new TaskD();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskD {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            String s = in.next();
            int[] d1 = new int[s.length()];
            int l = 0, r = -1;
            for (int i = 0; i < s.length(); ++i) {
                int k = i > r ? 1 : Math.min(d1[l + r - i], r - i);
                while (i + k < s.length() && 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;
                }
            }
            int[] d2 = new int[s.length()];
            l = 0;
            r = -1;
            for (int i = 0; i < s.length(); ++i) {
                int k = i > r ? 1 : Math.min(d2[l + r - i + 1], r - i + 1);
                while (i + k - 1 < s.length() && i - k >= 0 && s.charAt(i + k - 1) == s.charAt(i - k)) ++k;
                d2[i] = Math.max(--k, 0);
                if (i + k - 1 > r) {
                    l = i - k;
                    r = i + k - 1;
                }
            }
            int[] lDer = new int[s.length() + 1];
            int[] rDer = new int[s.length() + 1];
            int[] left = new int[s.length()];
            int[] right = new int[s.length()];
            int integ1 = 0;
            int integ2 = 0;
            for (int i = 0; i < d1.length; ++i) {
                lDer[i]++;
                lDer[i + d1[i]]--;
                rDer[i + 1]++;
                rDer[i - d1[i] + 1]--;
                if (d2[i] != 0) {
                    lDer[i]++;
                    lDer[i + d2[i]]--;
                    rDer[i]++;
                    rDer[i - d2[i]]--;
                }
            }
            for (int i = 0; i < d1.length; ++i) {
                integ1 += lDer[i];
                integ2 += integ1;
                left[i] = integ2;
            }
            integ1 = 0;
            for (int i = d1.length; i > 0; i--) {
                integ1 += rDer[i];
                right[i - 1] = integ1;
            }
            long ans = 0;
            for (int i = 0; i < d1.length - 1; ++i) {
                ans += left[i] * right[i + 1];
            }
            out.println(ans);
        }

    }

    static class InputReader {
        private StringTokenizer tokenizer;
        private BufferedReader reader;

        public InputReader(InputStream inputStream) {
            reader = new BufferedReader(new InputStreamReader(inputStream));
        }

        private void fillTokenizer() {
            if (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (Exception e) {
                    throw new RuntimeException(e);
                }
            }
        }

        public String next() {
            fillTokenizer();
            return tokenizer.nextToken();
        }

    }
}