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.*;
import java.util.*;

public class Solution {


	private StringTokenizer st;
	private BufferedReader in;
	private PrintWriter out;

	long solve(long maxu, long maxv, long maxsum) {
		maxu = Math.min(maxu, maxsum);
		maxv = Math.min(maxv, maxsum);
		long k = (maxu + maxv - 2) - (maxsum - 1);
		if (k < 0) {
			k = 0;
		}
		return maxu * maxv - k * (k + 1) / 2;
	}
	
	public void solve() throws IOException {
		String s = next();
		int n = s.length();
		int[] c1 = new int[n]; // i-u .. i+u
		for (int i = 0; i < n; ++i) {
			while (i - c1[i] >= 0 && i + c1[i] < n && s.charAt(i - c1[i]) == s.charAt(i + c1[i])) {
				++c1[i];
			}
		}
		int[] c2 = new int[n - 1]; // i-v .. i+v+1
		for (int i = 0; i < n - 1; ++i) {
			while (i - c2[i] >= 0 && i + c2[i] + 1 < n && s.charAt(i - c2[i]) == s.charAt(i + c2[i] + 1)) {
				++c2[i];
			}
		}
		//System.out.println(Arrays.toString(c1) + " " + Arrays.toString(c2));
		long ans = 0;
		for (int i = 0; i < n; ++i) {
			for (int j = i + 1; j < n; ++j) {
				//u+v<j-i
				ans += solve(c1[i], c1[j], j - i);
				if (j < n - 1) {
					ans += solve(c1[i], c2[j], j - i);
				}
			}
		}
		for (int i = 0; i < n - 1; ++i) {
			for (int j = i + 2; j < n; ++j) {
				//u+v<j-i-1
				ans += solve(c2[i], c1[j], j - i - 1);
				if (j < n - 1) {
					ans += solve(c2[i], c2[j], j - i - 1);
				}
			}
		}
		out.println(ans);
	}

	public void run() throws IOException {
		in = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		eat("");
		
		solve();
		
		out.close();
	}
	
	void eat(String s) {
		st = new StringTokenizer(s);
	}
	
	String next() throws IOException {
		while (!st.hasMoreTokens()) {
			String line = in.readLine();
			if (line == null) {
				return null;
			}
			eat(line);
		}
		return st.nextToken();
	}
	
	int nextInt() throws IOException {
		return Integer.parseInt(next());
	}
	
	long nextLong() throws IOException {
		return Long.parseLong(next());
	}
	
	double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}
	
	public static void main(String[] args) throws IOException {
		Locale.setDefault(Locale.US);
		new Solution().run();
	}
	
}