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 PalindromePairs {
    public static int n;
    public static int[] a;
    public static int[] b;
    public static void main(String[] args) throws IOException {
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        String s = f.readLine();
        n = s.length();
        a = new int[n];
        b = new int[n];
        
        for (int i = 0; i < n; i++) {
            int l = i;
            int r = i;
            add(l,r);
            while(l-1 >= 0 && r+1 < n && s.charAt(l-1) == s.charAt(r+1))
                add(--l, ++r);
        }
        
        for (int i = 0; i < n-1; i++) 
            if (s.charAt(i) == s.charAt(i+1)) {
                int l = i;
                int r = i+1;
                add(l,r);
                while(l-1 >= 0 && r+1 < n && s.charAt(l-1) == s.charAt(r+1))
                    add(--l, ++r);
            }
        
        for (int i = 1; i < n; i++)
            a[i] += a[i-1];
        long sum = 0;
        for (int i = 0; i < n-1; i++)
            sum += 1l*a[i]*b[i+1];
        System.out.println(sum);
    }
    
    public static void add(int l, int r) {
        a[r]++;
        b[l]++;
    }
}