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.*;
import java.util.LinkedList;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.math.BigInteger;

public class tmp {
    public static void main(String [] args) throws Exception{
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
  //      StringTokenizer st = new StringTokenizer(in.readLine());
        StringBuilder sb = new StringBuilder();
        
        String tmp = in.readLine();
        char[] s = tmp.toCharArray();
        int i,j,k;
        long count=0;
        int[] front=new int[2000],rear=new int[2000];

        for(i=0;i<s.length;i++){
                for(j=i,k=i;j>=0&&k<s.length&&s[k]==s[j];j--,k++){
                        front[j]++;rear[k]++;
                }
                for(j=i,k=i+1;j>=0&&k<s.length&&s[k]==s[j];j--,k++){
                        front[j]++;rear[k]++;
                }
        }
        for(i=s.length-1;i>0;i--)
                front[i-1]+=front[i];
        for(i=0;i<s.length-1;i++)
                count+=rear[i]*front[i+1];
        System.out.println(count);
        
        
        
        
        
        
        
        
        
        
        
        
        
        
    }
    
    public static long sum(int[][] arr, int start, int end){
        long out = 0;
        for(int i=start; i<=end; i++){
            out+=arr[i][end];
        }
        return out;
    }
    
    public static int isPalindrome(String s){
        for(int i=0; i<=s.length()/2; i++){
            if(s.charAt(i) != s.charAt(s.length()-1-i))
                return 0;
        }
        return 1;
    }
    
    
    
    
    
    
    
    


    
    
    
    
}