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.*;
import java.util.StringTokenizer;

public class TaskD extends Thread {
    public TaskD(String inputFileName, String outputFileName) {
        try {
            if (inputFileName != null) {
                this.input = new BufferedReader(new FileReader(inputFileName));
            } else {
                this.input = new BufferedReader(new InputStreamReader(System.in));
            }
            if (outputFileName != null) {
                this.output = new PrintWriter(outputFileName);
            } else {
                this.output = new PrintWriter(System.out);
            }
            this.setPriority(Thread.MAX_PRIORITY);
        } catch (Throwable e) {
            System.err.println(e.getMessage());
            e.printStackTrace();
            System.exit(666);
        }
    }

    private void solve() throws Throwable {
        String s = next();
        int n = s.length();
        int endHere[] = new int[n];
        int beginHere[] = new int[n];
        for (int pos = 0; pos < n; ++pos) {
            ++endHere[pos];
            ++beginHere[pos];
            int l = pos - 1, r = pos + 1;
            while (l >= 0 && r < n && s.charAt(l) == s.charAt(r)) {
                ++beginHere[l];
                ++endHere[r];
                --l;
                ++r;
            }
            l = pos;
            r = pos + 1;
            while (l >= 0 && r < n && s.charAt(l) == s.charAt(r)) {
                ++beginHere[l];
                ++endHere[r];
                --l;
                ++r;
            }
        }
        for (int i = n - 2; i >= 0; --i) {
            beginHere[i] += beginHere[i + 1];
        }
        long ret = 0;
        for (int i = 0; i + 1 < n; ++i) {
            ret += endHere[i] * beginHere[i + 1];
        }
        output.println(ret);
    }

    public void run() {
        try {
            solve();
        } catch (Throwable e) {
            System.err.println(e.getMessage());
            e.printStackTrace();
            System.exit(666);
        } finally {
            output.close();
        }
    }

    public static void main(String... args) {
        new TaskD(null, null).start();
    }

    private int nextInt() throws IOException {
        return Integer.parseInt(next());
    }

    private double nextDouble() throws IOException {
        return Double.parseDouble(next());
    }

    private long nextLong() throws IOException {
        return Long.parseLong(next());
    }

    private String next() throws IOException {
        while (tokens == null || !tokens.hasMoreTokens()) {
            tokens = new StringTokenizer(input.readLine());
        }
        return tokens.nextToken();
    }

    private StringTokenizer tokens;
    private BufferedReader input;
    private PrintWriter output;
}