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.util.ArrayList;
import java.util.HashMap;
import java.util.HashSet;

public class MainD {
    private static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    private static String  str;
    private static char[] s;
    private static boolean [][] ispal;


    private static boolean pal(int a, int b, int size ){
        if (a > 0 && b < size - 1 && ispal[a-1][b+1] && s[a] == s[b])
            return true;
        for (int i = 0; i != (b - a + 1)/2; ++i){
            if (s[a+i] != s[b - i]){
                return false;
            }
        }
        return true;
    }



    public static void main(String[] args) throws IOException {
        str = reader.readLine();
        s = str.toCharArray();
        int size = str.length();
        long[] a = new long[size];
        long[] b = new long[size];
        ispal = new boolean[size][size];
        for (int i = 0; i != size; ++i){
            a[i] = 0;
            b[i]=0;
        }
        for (int i = 0; i != size; ++i){
            for (int j = i; j != size; ++j){
            ispal[i][j] = false;
        }}
        long counter = 0;
        for (int i = 0; i != size; ++i){
            for (int j = i; j != size; ++j){
                if (pal(i,j,size)){
                    ispal[i][j] = true;
                    b[j]++;

                    if(j == 0){

                       a[j] = b[j];
                    }else{

                        a[j] = b[j] + a[j -1];
                    }
                    if (i != 0)
                       counter += a[i - 1];
                }
            }
        }


        System.out.print(counter);
    }

}