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.InputStreamReader;
import java.io.IOException;
import java.awt.Point;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.math.BigInteger;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author Maxim Gladkikh 
 */
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) {
        long mul = (int) 1e9 + 7;
        char[] s = in.next().toCharArray();
        int n = s.length;
        long[] powers = new long[n + 1];
        powers[0] = 1;
        long[] hashes = new long[n + 1];
        long[] revHashes = new long[n + 1];
        for (int i = 0; i < n; ++i) {
            powers[i + 1] = powers[i] * mul;
            hashes[i + 1] = hashes[i] * mul + s[i];
            revHashes[i + 1] = revHashes[i] * mul + s[n - 1 - i];
        }
        int[] start = new int[n];
        int[] finish = new int[n];
        for (int len = 1; len <= n; ++len) {
            for (int pos = 0; pos + len <= n; ++pos) {
                int revPos = n - pos - len;
                long h1 = hashes[pos + len] - hashes[pos] * powers[len];
                long h2 = revHashes[revPos + len] - revHashes[revPos] * powers[len];
                if (h1 == h2) {
                    ++start[pos];
                    ++finish[pos + len - 1];
                }
            }
        }
        long ans = 0;
        for (int i = 0; i < n; ++i) {
            int finished = 0;
            finished += finish[i];
            for (int j = i + 1; j < n; ++j) {
                ans += 1l * finished * start[j];
            }
        }
        out.println(ans);
    }
}

class InputReader {
    private StringTokenizer tokenizer;
    private BufferedReader reader;

    public InputReader(InputStream inputStream) {
        reader = new BufferedReader(new InputStreamReader(inputStream));
    }

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

    }