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.*;

public class Main {
    public static void main (String [] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
//        StringTokenizer st = new StringTokenizer(br.readLine());
        char[] s = br.readLine().toCharArray();
        int slen = s.length;
        long npoly = 0;
        //p1 and p2 lens = 1
        npoly += (slen * (slen - 1)) / 2;

        //find all poly with len > 1
        int[] polyc1 = new int[slen];
        int[] polyc2 = new int[slen];
        int[] a = new int[slen];
        int[] b = new int[slen];
        // poly len % 2 = 1
        int l = 0, r = -1, k;
        for (int i = 0; i < slen; ++i) {
            if (i > r) {
                k = 1;
            } else {
                k = Math.min(polyc1[l + r - i], r - i) + 1;
            }
            while (i + k < slen && i - k >= 0 && s[i + k] == s[i - k]) {
                ++k;
            }
            --k;
            polyc1[i] = k;
            //p1 and p2 len1 = 1 len2 > 1
            for (int j = 1; j <= k; ++j) {
                npoly += slen - (j * 2 + 1);
                ++a[i - j]; ++b[i + j];
            }
            if (i + k > r) {
                l = i - k;  r = i + k;
            }
        }
        // poly len % 2 = 0
        l = 0; r = -1;
        for (int i = 0; i < slen; ++i) {
            if (i > r) {
                k = 1;
            } else {
                k = Math.min(polyc2[l + r - i + 1], r - i + 1) + 1;
            }
            while (i + k - 1 < slen && i - k >= 0 && s[i + k - 1] == s[i - k]) {
                ++k;
            }
            --k;
            polyc1[i] = k;
            //p1 and p2 len1 = 1 len2 > 1
            for (int j = 1; j <= k; ++j) {
                npoly += slen - (j * 2);
                ++a[i - j]; ++b[i + j - 1];
            }
            if (i + k - 1 > r) {
                l = i - k;  r = i + k - 1;
            }
        }

        //p1 and p2 lens > 1
        for (int i = 1; i < slen - 2; ++i) {
            for (int j = i + 1; j < slen - 1; ++j) {
                npoly += b[i] * a[j];
            }
        }

        System.out.println(npoly);
    }
}