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

public class PalindromePairs {

	static char[] c;
	static int n;
	static int[][] dp;

	static int solve(int i, int j) {
		if (i == j)
			return 1;
		if (i > j)
			return 0;
		if (dp[i][j] != -1)
			return dp[i][j];
		if (c[i] == c[j])
			return dp[i][j] = 2 + solve(i + 1, j - 1);
		return dp[i][j] = Math.max(solve(i + 1, j), solve(i, j - 1));
	}

	public static void main(String[] args) throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(System.out);
		sc = new StringTokenizer(br.readLine());
		c = nxtCharArr();
		n = c.length;
		dp = new int[n][n];
		for (int i = 0; i < n; i++)
			Arrays.fill(dp[i], -1);
		int[] cnt = new int[n];
		for (int i = 0; i < n; i++)
			for (int j = i; j < n; j++)
				if (solve(i, j) == j - i + 1)
					cnt[i]++;
		for (int i = 1; i < n; i++)
			cnt[i] += cnt[i - 1];
		long ans = 0;
		for (int i = 0; i < n; i++)
			for (int j = i; j < n; j++)
				if (solve(i, j) == j - i + 1)
					ans += cnt[n - 1] - cnt[j];
		out.println(ans);
		br.close();
		out.close();
	}

	static BufferedReader br;

	static StringTokenizer sc;

	static String nxtTok() throws IOException {
		while (!sc.hasMoreTokens()) {
			String s = br.readLine();
			if (s == null)
				return null;
			sc = new StringTokenizer(s.trim());
		}
		return sc.nextToken();
	}

	static int nxtInt() throws IOException {
		return Integer.parseInt(nxtTok());
	}

	static long nxtLng() throws IOException {
		return Long.parseLong(nxtTok());
	}

	static double nxtDbl() throws IOException {
		return Double.parseDouble(nxtTok());
	}

	static int[] nxtIntArr(int n) throws IOException {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = nxtInt();
		return a;
	}

	static long[] nxtLngArr(int n) throws IOException {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nxtLng();
		return a;
	}

	static char[] nxtCharArr() throws IOException {
		return nxtTok().toCharArray();
	}
}