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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;

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

    static class TaskD {
        public void solve(int testNumber, FastReader in, PrintWriter out) {
            char s[] = in.nextString().toCharArray();
            TaskD.Manacher man = new TaskD.Manacher(s);

            int n = s.length;
            int start[] = new int[n + 1];
            int end[] = new int[n + 1];

            for (int i = 0, j = 0; i < man.p.length; i += 2, j++) {
                start[j + 1]--;
                start[j - man.p[i] / 2]++;
                end[j]++;
                end[j + man.p[i] / 2 + 1]--;
            }

            for (int i = 1, j = 0; i < man.p.length; i += 2, j++) {
                start[j + 1]--;
                start[j - man.p[i] / 2 + 1]++;
                end[j + 1]++;
                end[j + 1 + man.p[i] / 2]--;
            }

            for (int i = 1; i < n; i++) {
                start[i] += start[i - 1];
                end[i] += end[i - 1];
            }

            for (int i = n - 2; i >= 0; i--)
                start[i] += start[i + 1];

            long ans = 0;
            for (int i = 0; i < n - 1; i++)
                ans += end[i] * start[i + 1];

            out.println(ans);
        }

        static public class Manacher {
            private int[] p;
            private char[] s;
            private char[] t;

            public Manacher(char[] s) {
                this.s = s;
                preprocess();
                p = new int[t.length];

                int center = 0, right = 0;
                for (int i = 1; i < t.length - 1; i++) {
                    int mirror = 2 * center - i;

                    if (right > i)
                        p[i] = Math.min(right - i, p[mirror]);

                    while (t[i + (1 + p[i])] == t[i - (1 + p[i])])
                        p[i]++;

                    if (i + p[i] > right) {
                        center = i;
                        right = i + p[i];
                    }
                }

                int ans[] = new int[p.length - 4];
                for (int i = 2; i < p.length - 2; i++)
                    ans[i - 2] = p[i];

                p = ans;
            }

            private void preprocess() {
                t = new char[s.length * 2 + 3];
                t[0] = '$';
                t[s.length * 2 + 2] = '@';
                for (int i = 0; i < s.length; i++) {
                    t[2 * i + 1] = '#';
                    t[2 * i + 2] = s[i];
                }
                t[s.length * 2 + 1] = '#';
            }

        }

    }

    static class FastReader {
        private InputStream stream;
        private byte[] buf = new byte[8192];
        private int curChar;
        private int pnumChars;

        public FastReader(InputStream stream) {
            this.stream = stream;
        }

        private int pread() {
            if (pnumChars == -1) {
                throw new InputMismatchException();
            }
            if (curChar >= pnumChars) {
                curChar = 0;
                try {
                    pnumChars = stream.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (pnumChars <= 0) {
                    return -1;
                }
            }
            return buf[curChar++];
        }

        public String nextString() {
            int c = pread();
            while (isSpaceChar(c))
                c = pread();
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = pread();
            } while (!isSpaceChar(c));
            return res.toString();
        }

        private boolean isSpaceChar(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

    }
}