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.ArrayList;
import java.util.Scanner;

public class VKChgRound2ProbD {

    static char[] a;
    static int N;
    static ArrayList<Integer> g[];
    static int elements[];

    public static void calc(int i, int j) {
        if (i > -1 && j < N && a[i] == a[j]) {
            g[i].add(j);
            elements[i]++;
            calc(i - 1, j + 1);
        }
    }

    static int[] allE;

    public static void genAllElms() {
        allE = new int[N + 1];

        for (int i = N - 1; i > -1; i--) {
            allE[i] = allE[i + 1] + elements[i];
        }
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);

        a = in.next().toCharArray();
        N = a.length;
        g = new ArrayList[N + 1];
        elements = new int[N];
        for (int i = 0; i < N; i++)
            g[i] = new ArrayList<Integer>();
        calc(0, 0);
        for (int i = 1; i < N; i++) {
            calc(i, i);
            calc(i - 1, i);
        }

        genAllElms();

        long ans = 0;
        for (int i = 0; i < N; i++) {
            for (int limit : g[i]) {
                ans += allE[limit+1];
            }
        }
        System.out.println(ans);
    }

}