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.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Set;


public class Task4 {

	/**
	 * @param args
	 * @throws IOException 
	 */
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String s = br.readLine();
        int len = s.length();
        char[] cs = new char[len];
        s.getChars(0, len, cs, 0);
        long[] p = new long[len];
        Map<Integer, Set<Integer>> m = new HashMap<Integer, Set<Integer>>(); 
        long sum = 0;
        p[0] = 1;
        m.put(0, new HashSet<Integer>(5));
        Set<Integer> set = m.get(0);
        set.add(0);
        for (int i = 1; i < len; i++) {
        	Set<Integer> newSet = new HashSet<Integer>(5);
        	newSet.add(i);
        	sum += p[i-1];
        	if (cs[i-1] == cs[i]) {
        		newSet.add(i-1);
        		if (i > 1) sum += p[i-2];
        	}
        	for(int j : set) {
        		if (j > 0 && cs[j-1] == cs[i]) {
        			newSet.add(j-1);
        			if (j > 1) sum += p[j-2];
        		}
        	}
        	m.put(i, newSet);
        	p[i] = p[i-1] + newSet.size();
        	set = newSet;
        }
        System.out.println(sum);
	}
}