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.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) throws Exception{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String s = br.readLine();
        int n = s.length();
        int[][] arr = new int[n][n];
        for(int i=0;i<n;i++){
            arr[i][i]=1;
        }
        for(int i=0;i<n-1;i++){
            if(s.charAt(i)==s.charAt(i+1))arr[i][i+1]=1;
        }
        for(int len=3;len<=s.length();len++){
            for(int start=0;start+len-1<s.length();start++){
                int end=start+len-1;
                if(s.charAt(start)==s.charAt(end))arr[start][end]=arr[start+1][end-1];
            }
        }
        int[] count = new int[n+5];
        for(int i=0;i<n;i++){
            for(int j=i;j<n;j++){
                if(arr[i][j]==1)count[j]++;
            }
        }
        for(int i=1;i<n;i++){
            count[i]+=count[i-1];
        }
        long ans = 0l;
        for(int i=0;i<n;i++){
            int first = count[i];
            int second = 0;
            for(int j=i+1;j<n;j++){
                if(arr[i+1][j]==1)second++;
            }
            ans+= first*second;
        }
        System.out.println(ans);
    }
   
}