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.TreeMap;

public class Main {
	public static void main(String[] args) throws IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));

		String s = in.readLine();
		
		char[] str = s.toCharArray();
		
		int size = s.length();
		
		int[] endIn = new int[size];
		int[] strIn = new int[size];
		
		long ans = 0;
	
		for (int i = 0 ; i < s.length() ; i++) {
			int b, e;
			
			b = i;
			e = i;
			
			while (b >= 0 && e < s.length() && str[b] == str[e]) {
				endIn[e]++;
				strIn[b]++;
				b--;
				e++;
			}
			
			
			b = i - 1;
			e = i;
				
			while (b >= 0 && e < s.length() && str[b] == str[e]) {
				endIn[e]++;
				strIn[b]++;
				b--;
				e++;
			}
		}
		
		for (int i = 1 ; i < size ; i++) {
			ans += (long) endIn[i - 1] * (long) strIn[i];
			endIn[i] += endIn[i - 1];
		}	
		
		System.out.println(ans);
	}
}