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.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class TaskD implements Runnable {
    private BufferedReader reader;
    private PrintWriter writer;
    private StringTokenizer tokenizer;

    public static void main(String[] args) {
        new Thread(new TaskD()).start();
    }

    private String next() throws Exception {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            tokenizer = new StringTokenizer(reader.readLine());
        }

        return tokenizer.nextToken();
    }

    @Override
    public void run() {
        try {
            reader = new BufferedReader(new InputStreamReader(System.in));
            writer = new PrintWriter(System.out);

            solve();

            reader.close();
            writer.close();
        } catch (Exception e) {
            e.printStackTrace();
            System.exit(1);
        }
    }

    private String s;
    private int n;

    private long[] cntB;
    private long[] cntE;

    private void solve() throws Exception {
        s = next();
        n = s.length();

        countPalindromes();

        long result = 0;

        for (int i = 0; i < n - 1; i++) {
            for (int j = i + 1; j < n; j++) {
                result += cntE[i] * cntB[j];
            }
        }

        writer.println(result);
    }

    private void countPalindromes() {
        int[] d1 = findD1();
        int[] d2 = findD2();

        cntB = new long[n];
        cntE = new long[n];

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < d1[i]; j++) {
                cntB[i - j]++;
                cntE[i + j]++;
            }

            for (int j = 0; j < d2[i]; j++) {
                cntB[i - j - 1]++;
                cntE[i + j]++;
            }
        }
    }

    private int[] findD1() {
        int[] d1 = new int[n];

        int l = 0;
        int r = -1;

        for (int i = 0; i < n; i++) {
            int k = (i > r ? 0 : Math.min(d1[l + r - i], r - i)) + 1;

            while (i + k < n && i - k >= 0 && s.charAt(i + k) == s.charAt(i - k)) {
                k++;
            }

            d1[i] = --k;

            if (i + k > r) {
                l = i - k;
                r = i + k;
            }
        }

        for (int i = 0; i < n; i++) {
            d1[i]++;
        }

        return d1;
    }

    private int[] findD2() {
        int[] d2 = new int[n];

        int l = 0;
        int r = -1;

        for (int i = 0; i < n; i++) {
            int k = (i > r ? 0 : Math.min(d2[l + r - i + 1], r - i + 1)) + 1;

            while (i + k - 1 < n && i - k >= 0 && s.charAt(i + k - 1) == s.charAt(i - k)) {
                k++;
            }

            d2[i] = --k;

            if (i + k - 1 > r) {
                l = i - k;
                r = i + k - 1;
            }
        }

        return d2;
    }
}