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.*;

public class d{
    public static void main(String args[]){
	Scanner sc = new Scanner(System.in);
	char[] s = sc.next().toCharArray();
	int len = s.length;

	int[] a = new int[len];
	int[] b = new int[len];

	for(int i=0; i<len; i++){
	    for(int j=0; i+j<len && i-j>=0; j++){
		if(s[i+j] != s[i-j])
		    break;
		a[i+j]++;
		b[i-j]++;
	    }
	}

	for(int i=1; i<len; i++){
	    for(int j=1; i+j-1<len && i-j>=0; j++){
		if(s[i+j-1] != s[i-j])
		    break;
		a[i+j-1]++;
		b[i-j]++;
	    }
	}
	
	long ans = 0;
	for(int i=0; i<len-1; i++){
	    for(int j=i+1; j<len; j++){
		ans += a[i]*b[j];
	    }
	}

	System.out.println(ans);

    }
}