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.util.Arrays;

public class D {
    static int[][] dp;
    static String str;

    public static int solve(int i, int j) {
        if (i == j) {
            return 1;
        }
        if (str.charAt(i) == str.charAt(j)) {
            if (j == i + 1) {
                return 1;
            } else {
                if (dp[i][j] != -1)
                    return dp[i][j];
                if (solve(i + 1, j - 1) == 1)
                    return dp[i][j] = 1;
                return dp[i][j] = 0;
            }
        } else {
            return dp[i][j] = 0;
        }
    }

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        str = br.readLine();
        dp = new int[str.length()][str.length()];
        for (int i = 0; i < str.length(); i++)
            Arrays.fill(dp[i], -1);
        long[] endingWith = new long[str.length()];
        long[] startingWith = new long[str.length()];
        for (int i = 0; i < str.length(); i++) {
            for (int j = i; j < str.length(); j++) {
                if (solve(i, j) == 1) {
                    startingWith[i]++;
                    endingWith[j]++;
                }
            }
        }
        long[] sumStarting = new long[str.length()];
        sumStarting[str.length() - 1] = startingWith[str.length() - 1];
        for (int i = str.length() - 2; i >= 0; i--) {
            sumStarting[i] = sumStarting[i + 1] + startingWith[i];
        }
        long count = 0;
        for (int i = 0; i < str.length() - 1; i++) {
            count += endingWith[i] * sumStarting[i + 1];
        }
        System.out.println(count);
    }
}