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

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

class TaskD {
    static boolean t[][];
    static int start[];
    static int dp[];

    public void solve(int testNumber, Scanner in, PrintWriter out) {
        String s = in.next();
        char[] st = s.toCharArray();
        BuildPalindrome(st);
        out.print(CalculatePairs(st));
    }

    private long CalculatePairs(char[] s) {
        int n = s.length;
        long ans = 0;
        for (int i = 0; i < n; i++) {
            for (int j = i; j < n; j++) {
                if (t[i][j]) {
                    ans += dp[j+1];
                }
            }
        }
        return ans;
    }

    private void BuildPalindrome(char[] s) {
        int n = s.length;
        t = new boolean[n][n];
        start = new int[n];
        dp =new int[n+1];
        //1 length palindromes
        for (int i = 0; i < n; i++)
            t[i][i] = true;
        for (int i = 1; i < n; i++) {
            for (int j = 0; j < n-i ; j++) {
                int p = j,q=j+i;
                if (s[p] == s[q])
                {
                    //length 2 - true
                    if(i==1) t[j][i+j] = true;
                    else if(t[p+1][q-1]) t[p][q] = true;
                }
            }
        }
        //fill start array (no of palindromes that start with i)
        for (int i = 0; i < n; i++) {
            for (int j = i; j < n; j++) {
                if (t[i][j])
                    ++start[i];
            }
        }
        dp[n-1] = start[n-1];
        //store cumulative sum to reduce complexity.
        for (int i = n-2; i >= 0; i--) {
           dp[i] = dp[i+1] + start[i];
        }
    }
}