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.io.PrintWriter;
import java.util.StringTokenizer;

/**
 * Created by IntelliJ IDEA.
 * User: Den
 * Date: 09.03.12
 * Time: 17:07
 * To change this template use File | Settings | File Templates.
 */

public class D extends Thread {

    public void run() {
        try {
            String s = nextToken();
            char[] a = s.toCharArray();
            int n = s.length();
            int[] polEnds = new int[n];
            int[] polStarts = new int[n];

            for (int i = 0; i < n; ++i) {
                polEnds[i]++;
                polStarts[i]++;
                for (int j = 1; (i-j)>=0 && (i+j)<n && a[i+j]==a[i-j]; j++) {
                    polEnds[i+j]++;
                    polStarts[i-j]++;
                }
                for (int j=0;(i-j)>=0 && (i+j+1)<n && a[i+j+1]==a[i-j]; j++){
                    polEnds[i+j+1]++;
                    polStarts[i-j]++;
                }
            }

            for (int i=n-2; i>=0; i--){
                polStarts[i]+=polStarts[i+1];                
            }        
            
            long ans = 0L;
            for (int i=0; i<n-1; i++){
                ans+=(long)polEnds[i]*polStarts[i+1];
            }

            out.print(ans);
        } catch (Exception ex) {
            ex.printStackTrace();
            System.exit(999);
        } finally {
            out.flush();
            out.close();
        }
    }


    public static void main(String[] args) {
        new D().run();
    }

    public D() {
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        setPriority(Thread.MAX_PRIORITY);
    }

    private BufferedReader in;
    private PrintWriter out;
    private StringTokenizer stringTokenizer;

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

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

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

    private String nextToken() throws IOException {
        if (stringTokenizer == null || !stringTokenizer.hasMoreElements()) {
            stringTokenizer = new StringTokenizer(in.readLine(), "\t\n\r ");
        }
        return stringTokenizer.nextToken();
    }
}