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;

public class DD {
    static char[] a;
    static int N;
    static int[] tree;
    static int[][] dp;

    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        a = in.readLine().toCharArray();
        N = a.length + 10;
        tree = new int[N];
        dp = new int[a.length + 10][a.length + 10];
        int[] e = new int[2001999];
        int si = 0;
        long sum = 0;
        for (int i = 0; i < a.length; i++)
            for (int j = i; j < a.length; j++)
                dp[i][j] = -1;

        for (int i = 0; i < a.length; i++) {
            for (int j = i; j < a.length; j++) {
                if (isPal(i, j)) {
                    e[si++] = j;
                    update(i + 1, 1);
                    sum++;
                }
            }
        }
        long res = 0;
        for (int i = 0; i < si; i++)
            res += sum - read(e[i] + 1);
        System.out.println(res);
    }

    public static boolean isPal(int l, int r) {
        if (l >= r)
            return true;
        if (a[l] != a[r]) {
            dp[l][r] = 0;
            return false;
        }
        if (dp[l][r] != -1)
            return dp[l][r] == 1;
        boolean ok = isPal(l + 1, r - 1);
        dp[l][r] = ok ? 1 : 0;
        return ok;
    }

    public static void update(int idx, int val) {
        while (idx < N) {
            tree[idx] += val;
            idx += (idx & -idx);
        }
    }

    public static int read(int idx) {
        int sum = 0;
        while (idx > 0) {
            sum += tree[idx];
            idx -= (idx & -idx);
        }
        return sum;
    }
}