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 TaskD {
	BufferedReader br;
	PrintWriter out;
	StringTokenizer stok;

	String nextToken() throws IOException {
		while (stok == null || !stok.hasMoreTokens()) {
			String s = br.readLine();
			if (s == null) {
				return null;
			}
			stok = new StringTokenizer(s);
		}
		return stok.nextToken();
	}

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

	long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}

	char nextChar() throws IOException {
		return (char) (br.read());
	}

	String nextLine() throws IOException {
		return br.readLine();
	}

	final int MAXL = 2031;
	long[] hash = new long[MAXL];
	long[] hashBack = new long[MAXL];
	int n;

	long getHash(int l, int r) {
		if (l == 0) {
			return hash[r];
		}
		return hash[r] - hash[l - 1] * step[r - l + 1];
	}

	long getHashBack(int l, int r) {
		if (r == n - 1) {
			return hashBack[l];
		}
		return hashBack[l] - hashBack[r + 1] * step[r - l + 1];
	}

	long[] step = new long[MAXL];

	void solve() throws IOException {
		String s = nextToken();
		n = s.length();
		int p = 31;
		step[0] = 1;
		for (int i = 1; i < MAXL; i++) {
			step[i] = step[i - 1] * p;
		}
		hash[0] = (int) s.charAt(0);
		for (int i = 1; i < s.length(); i++) {
			hash[i] = hash[i - 1] * p + (int) s.charAt(i);
		}

		hashBack[n - 1] = (int) s.charAt(n - 1);
		for (int i = n - 2; i >= 0; i--) {
			hashBack[i] = hashBack[i + 1] * p + (int) s.charAt(i);
		}

		long[] start, finish;
		start = new long[n];
		finish = new long[n];
		for (int i = 0; i < n; i++) {
			for (int l = 1; l <= n - i; l++) {
				int len = (l + 1) / 2;
				if (l % 2 == 0) {
					if (getHash(i, i + len - 1) == getHashBack(i + len, i + 2
							* len - 1)) {
						start[i]++;
					}
					continue;
				}
				if (getHash(i, i + len - 1) == getHashBack(i + len - 1, i + 2
						* len - 2)) {
					start[i]++;
				}
			}
			for (int l = 1; l <= i + 1; l++) {
				int len = (l + 1) / 2;
				if (l % 2 == 0) {
					if (getHash(i - 2 * len + 1, i - len) == getHashBack(i
							- len + 1, i)) {
						finish[i]++;
					}
					continue;
				}
				if (getHash(i - 2 * len + 2, i - len + 1) == getHashBack(i
						- len + 1, i)) {
					finish[i]++;
				}
			}
		}
		long ans = 0;
		for (int i = 0; i < n; i++) {
			for (int j = i + 1; j < n; j++) {
				long tmp = finish[i] * start[j];
				ans += tmp;
			}
		}
		out.println(ans);
	}

	void run() throws IOException {
		// br = new BufferedReader(new FileReader("input.txt"));
		// out = new PrintWriter("output.txt");
		br = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		solve();
		br.close();
		out.close();
	}

	public static void main(String[] args) throws IOException {
		// Locale.setDefault(Locale.US);
		new TaskD().run();
	}
}