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;
public class Main {
    public static void main(String [] args) {
        Scanner read = new Scanner(System.in);
        String s = read.nextLine();
        int n = s.length();
        boolean [][] dp = new boolean [n][n];
        for(int i = 0; i < n; i ++) {
            dp[i][i] = true;
        }
        for(int i = 0; i < n - 1; i ++) {
            dp[i][i + 1] = (s.charAt(i) == s.charAt(i + 1));
        }
        for(int i = 3; i <= n; i ++) {
            for(int j = 0; j <= n - i; j ++) {
                dp[j][j + i - 1] = (dp[j + 1][j + i - 2] & s.charAt(j) == s.charAt(j + i - 1));
            }
        }
        int [] cnt = new int [n];
        for(int i = 1; i < n; i ++) {
            cnt[i] = 0;
            for(int j = i; j < n; j ++) {
                if(dp[i][j]) {
                    ++ cnt[i];
                }
            }
        }
        for(int i = n - 2; i >= 1; i --) {
            cnt[i] += cnt[i + 1];
        }
        long result = 0;
        for(int i = 0; i <= n - 2; i ++) {
            for(int j = i; j < n - 1; j ++) {
                if(dp[i][j]) {
                    result += cnt[j + 1];
                }
            }
        }
        System.out.println(result);
    }
}