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.util.List;
import java.io.IOException;
import java.util.Arrays;
import java.util.ArrayList;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.NoSuchElementException;
import java.util.TreeSet;
import java.util.Collection;
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;
        MyScanner in = new MyScanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskD solver = new TaskD();
        solver.solve(1, in, out);
        out.close();
    }
}

class TaskD {
    public void solve(int testNumber, MyScanner in, PrintWriter out) {
        String s = in.next();
        int n = s.length();
        StringBuilder sb = new StringBuilder();
        sb.append('$');
        for(char c:s.toCharArray()){
            sb.append(c);
            sb.append('$');
        }
        s = sb.toString();
        int[] rs = Palindrome.palindromeRadius(s);
        long[] left=new long[n],right=new long[n];
        for(int i=0;i<n;i++){
            for(int j=0;j<=2*i+1;j++){
                if(j + rs[j] >= 2*i+1){
                    left[i]++;
                }
            }
            for(int j=s.length()-1;j>=2*i+1;j--){
                if(j - rs[j] <= 2*i+1){
                    right[i]++;
                }
            }
        }
        long res = 0;
        for(int i=0;i<n;i++)for(int j=i+1;j<n;j++){
            res += left[i] * right[j];
        }
        out.println(res);
    }
}

class MyScanner {
    private final InputStream in;
    public MyScanner(InputStream in){
        this.in = in;
    }

    public String next() {
        try {
            StringBuilder res = new StringBuilder("");
            int c;
            do{
                c = in.read();
                if(c==-1)throw new NoSuchElementException();
            }while(Character.isWhitespace(c));
            do {
                res.append((char) c);
                c = in.read();
            } while (c!=-1 && !Character.isWhitespace(c));
            return res.toString();
        } catch (IOException e) {
            throw new NoSuchElementException();
        }
    }

    }

class Palindrome{
    public static int[] palindromeRadius(String s) {
        int n=s.length();
        int[] rad=new int[n];
        for(int i=0,j=0;i<n;) {
            while(j<n && i+i-j>=0 && s.charAt(j)==s.charAt(i+i-j))j++;
            rad[i]=j-i-1;
            for(int bi=i++;i<j && rad[bi+bi-i]!=rad[bi]+bi-i;i++) {
                rad[i] = Math.min(rad[bi]+bi-i,rad[bi+bi-i]);
            }
        }
        return rad;
    }

    }