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.InputStream;
import java.io.InputStreamReader;


public class Main {
	static final int size = 1000;
	
	BufferedReader reader;
	String row;
	long[] starts;
	long[] ends;
	int len;
	
	public Main(InputStream in) throws IOException {
		reader = new BufferedReader(new InputStreamReader(in));
		readParams();
	}
	/**
	 * @param args
	 * @throws IOException 
	 */
	public static void main(String[] args) throws IOException {
		Main m = new Main(System.in);
		System.out.print(m.solve());
	}
	
	public String solve() {
		len = row.length();
		starts = new long[len];
		ends = new long[len];
		findPolindromsN();
		for(int i = len - 2; i >= 0; i--) {
			starts[i] = starts[i] +  starts[i + 1];
		}
		long sum = 0;
		for(int i = 0; i < len - 1; i++) {
			sum += ends[i] * starts[i + 1];
		}
		return String.valueOf(sum);
	}
	
	private void findPolindromsN() {
		for(int i = 0; i < len; i ++) {
			int k = 0;
			while (i+k < len && i-k >= 0 && row.charAt(i+k) == row.charAt(i-k)) {
				starts[i-k]++;
				ends[i+k]++;
				k++;
			}
			k = 1;
			while (i+k-1 < len && i-k >= 0 && row.charAt(i+k-1) == row.charAt(i-k)) {
				starts[i-k]++;
				ends[i+k-1]++;
				k++;
			}
		}
		
	}
	public void readParams() throws IOException {
		
		row = readStr();
		
	}
	
	public Integer readInt() throws NumberFormatException, IOException {
		return Integer.parseInt(reader.readLine());
	}
	
	public Integer[] readInt(int size) throws IOException {
		Integer[] result = new Integer[size];
		String[] params =  reader.readLine().split(" ");
		for(int i = 0; i < size; i++) {
			result[i] = Integer.parseInt(params[i]);
		}
		return result;
	}
	
	public String readStr() throws IOException {
		return reader.readLine().trim();
	}

}