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 org.omg.stub.java.rmi._Remote_Stub;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class Solution implements Runnable {
    public static void main(String[] args) {
        (new Thread(new Solution())).start();
    }


    BufferedReader cin;
    PrintWriter cout;
    StringTokenizer tok;

    public String next() throws IOException {
        while (tok == null || !tok.hasMoreTokens()) {
            tok = new StringTokenizer(cin.readLine());
        }
        return tok.nextToken();
    }

    public int nextInt() throws IOException {
        return Integer.parseInt(next());
    }

    private void solve() throws IOException {
        cin = new BufferedReader(new InputStreamReader(System.in));
        cout = new PrintWriter(System.out);
        String s = next();
        int n = s.length();
        boolean[][] b = new boolean[n][n];
        for (int i = 0; i < n; i++) {
            int lf = i;
            int rg = i;
            while (lf >= 0 && rg < n) {
                if (s.charAt(lf) == s.charAt(rg)) {
                    b[lf][rg] = true;
                    --lf;
                    ++rg;
                } else {
                    break;
                }
            }
        }
        for (int i = 0; i < n - 1; i++) {
            int lf = i;
            int rg = i + 1;
            while (lf >= 0 && rg < n) {
                if (s.charAt(lf) == s.charAt(rg)) {
                    b[lf][rg] = true;
                    --lf;
                    ++rg;
                } else {
                    break;
                }
            }
        }
        int[] le = new int[n], re = new int[n];
        for (int i = 0; i < n; i++) {
            for (int j = i; j < n; j++) {
                if (b[i][j]) {
                    le[i]++;
                    re[j]++;
                }
            }
        }
        long ans = 0;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                ans += (long)le[j] * (long)re[i];
            }
        }
        cout.println(ans);
        cin.close();
        cout.close();
    }

    public void run() {
        try {
            solve();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}