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.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;

public class D {

        public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(System.out);
        String s = br.readLine();
        long ans = 0;
        int n = s.length();
        int[][] p = new int[n][n];
        int[] count = new int[n];
        
        for (int i = 0; i < n; i++) {
            int j = 0;
            while (i - j > -1 && i + j < n) {
                if (s.charAt(i - j) == s.charAt(i + j)) {
                    p[i - j][i + j] = 1;
                    count[i - j]++;
                } else
                    break;
                j++;
            }

            if (i < n - 1 && s.charAt(i) == s.charAt(i + 1)) {
                j = 0;
                while (i - j > -1 && i + j + 1 < n) {
                    if (s.charAt(i - j) == s.charAt(i + j + 1)) {
                        p[i - j][i + j + 1] = 1;
                        count[i - j]++;
                    } else
                        break;
                    j++;
                }
            }

        }

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

        for (int i = 0; i < n; i++)
            for (int j = 0; j < n - 1; j++) {
                if (p[i][j] == 1)
                    ans += count[j + 1];
            }

        pw.print(ans);
        pw.close();
    }

}