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.InputStream;
import java.io.InputStreamReader;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.IOException;
import java.util.StringTokenizer;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskD solver = new TaskD();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskD {
    public void solve(int testNumber, InputReader in, PrintWriter out) {
        char[] a = in.next().toCharArray();
        boolean[][] isPalindrome = new boolean[a.length][a.length];
        for (int i = a.length - 1; i >= 0; i--) {
            for (int j = i; j < a.length; j++) {
                if ((i + 1 > j - 1 || isPalindrome[i + 1][j - 1]) && a[i] == a[j]) {
                    isPalindrome[i][j] = true;
                }
            }
        }
        long[] f = new long[a.length];
        for (int i = 0; i < a.length; i++) {
            if (i > 0) f[i] += f[i - 1];
            for (int j = 0; j <= i; j++) {
                if (isPalindrome[j][i]) ++f[i];
            }
        }
        long res = 0;
        for (int i = 0; i + 1 < a.length; i++) {
            int count = 0;
            for (int j = i + 1; j < a.length; j++) {
                if (isPalindrome[i + 1][j]) ++count;
            }
            res += count * f[i];
        }
        out.println(res);
    }
}

class InputReader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;

    public InputReader(InputStream stream) {
        reader = new BufferedReader(new InputStreamReader(stream), 32768);
        tokenizer = null;
    }

    public String next() {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                tokenizer = new StringTokenizer(reader.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken();
    }

}