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 PalindromePairs {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);

        char[] x = br.readLine().toCharArray();
        int l = x.length;
        boolean[][] isPalin = new boolean[l][l];
        int[][] count = new int[l + 1][l + 1];

        for (int i = l - 1; i > -1; i--) {
            for (int j = i; j < l; j++) {
                if (i == j)
                    isPalin[i][j] = true;
                else if (j - i == 1)
                    isPalin[i][j] = x[i] == x[j];
                else {
                    if (isPalin[i + 1][j - 1])
                        isPalin[i][j] = x[i] == x[j];
                    else
                        isPalin[i][j] = false;
                }
            }
        }

        for (int i = l - 1; i > -1; i--) {
            for (int j = l - 1; j > -1; j--) {
                count[j][i] = count[j + 1][i];
                if (isPalin[j][i])
                    count[j][i]++;
            }
        }

        for (int i = l - 1; i > -1; i--) {
            for (int j = l - 1; j > -1; j--) {
                count[i][j] += count[i][j + 1];
            }
        }
        
        long answer = 0;
        for (int i = 0; i < l; i++) {
            for (int j = 0; j < l; j++) {
                if (isPalin[i][j]) {
                    answer += count[j + 1][j + 1];
                }
            }
        }

        out.println(answer);
        out.close();
    }
}