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 SolD implements Runnable {
	public static void main(String[] args) {
		new Thread(new SolD()).start();
	}

	BufferedReader br;
	StringTokenizer st;
	PrintWriter out;
	boolean eof;

	@Override
	public void run() {
		br = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		solve();
		out.close();
	}

	String nextToken() {
		while (st == null || !st.hasMoreTokens()) {
			try {
				st = new StringTokenizer(br.readLine());
			} catch (Exception e) {
				eof = true;
				return "0";
			}
		}
		return st.nextToken();
	}

	int nextInt() {
		return Integer.parseInt(nextToken());
	}

	long nextLong() {
		return Long.parseLong(nextToken());
	}
	
	void solve() {
		char[] ch = nextToken().toCharArray();
		int n = ch.length;
		int[] cnt = new int[n];
		boolean[][] pal = new boolean[n][n];
		for (int i = 0; i < n; i++) {
			int a = i;
			int b = i;
			while (a >= 0 && b < n && ch[a] == ch[b]) {
				cnt[a]++;
				pal[a][b] = true;
				a--;
				b++;
			}
		}
		for (int i = 0; i < n - 1; i++) {
			int a = i;
			int b = i + 1;
			while (a >= 0 && b < n && ch[a] == ch[b]) {
				cnt[a]++;
				pal[a][b] = true;
				a--;
				b++;
			}
		}
		int[] sum = new int[n + 1];
		for (int i = n - 1; i >= 0; i--) {
			sum[i] = sum[i + 1] + cnt[i];
		}
		long ans = 0;
		for (int i = 0; i < n; i++) {
			for (int j = i; j < n; j++) {
				if (pal[i][j]) {
					ans += sum[j + 1];
				}
			}
		}
		out.println(ans);
	}
}