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:

//package codeforces;

import java.util.Scanner;

public class D_159 {

    public void solve(){
        Scanner sc = new Scanner(System.in);
        String s = sc.next();
        boolean isPal[][] = new boolean[s.length()+1][s.length()+1];
        long[] dp = new long[s.length()+1];
        long[] p  = new long[s.length()+1];
        for(int i =1 ;  i<=s.length() ; i++){
            isPal[i][i] = true;
            for(int j=i-1; j>=1 ; j--){
                if(i-j==1){
                    if(s.charAt(i-1)==s.charAt(j-1)) isPal[j][i] = true;
                }else{
                    if(s.charAt(i-1)==s.charAt(j-1) && isPal[j+1][i-1]) isPal[j][i] = true;    
                }
            }
        }
        
        
        p[0] = p[1] = 1;
        for(int i=2; i <= s.length() ; i++){
            dp[i] = dp[i-1];
            p[i]  = p[i-1] + (isPal[1][i]?1:0);
            for(int j=i;j>1; j-- ){
                
                if(isPal[j][i]) dp[i] += p[j-1];
                p[i] += (isPal[j][i] ?  1 : 0 );
            }
            //System.out.println(p[i]);
        }
        
        System.out.println(dp[s.length()]);
        
        
        
    }
    
    public static void main(String args[]){
        new D_159().solve();
    }
}