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.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Solution {
	
	void solve() throws Exception {
		char[] s = nextToken().toCharArray();
		int n = s.length;
		int[] cend = new int[n];
		int[] cbegin = new int[n];
		for (int i = 0; i < n; i++) {
			for (int b = i, e = i; b >= 0 && e < n && s[b] == s[e]; b--, e++) {
				cbegin[b]++; cend[e]++;
			}
			for (int b = i - 1, e = i; b >= 0 && e < n && s[b] == s[e]; b--, e++) {
				cbegin[b]++; cend[e]++;
			}
		}
		long sumend = 0;
		long ans = 0;
		for (int i = 0; i < n - 1; i++) {
			sumend += cend[i];
			ans += sumend * cbegin[i + 1];
		}
		out.println(ans);
	}

	public static void main(String[] args) {
		new Solution().run();
	}
	
	public void run() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
			solve();
		} catch (Exception e) {
			e.printStackTrace();
			System.exit(1);
		} finally {
			out.close();
		}
	}
	
	String nextToken() throws Exception {
		while (st == null || !st.hasMoreTokens()) {
			String s = in.readLine();
			if (s == null) return null;
			st = new StringTokenizer(s);
		}
		return st.nextToken();
	}
	
	int nextInt() throws Exception {
		return Integer.parseInt(nextToken());
	}
	
	BufferedReader in;
	PrintWriter out;
	StringTokenizer st;

}