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

public class A {

	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);

		char[] s = sc.next().toCharArray();
		int n = s.length;

		boolean[][] isPal = new boolean[n][n];
		for(int i = 0; i < n; ++i)
			isPal[i][i] = true;
		for(int len = 2; len <= n; ++len)
			for(int i = 0; i + len <= n; ++i)
			{
				int j = i + len - 1;
				isPal[i][j] = s[i] == s[j] && (len == 2 || isPal[i + 1][j - 1]);
			}

		long[] countPals = new long[n + 1];
		for(int i = n - 1; i >= 0; --i)
		{
			countPals[i] = countPals[i + 1];
			for(int j = i; j < n; ++j)
				countPals[i] += isPal[i][j] ? 1 : 0;
		}
		long ans = 0;
		for(int len = 1; len <= n; ++len)
			for(int i = 0; i + len < n; ++i)
				if(isPal[i][i + len - 1])
					ans += countPals[i + len];
		out.println(ans);

		out.flush();
		out.close();
	}

	static class Scanner 
	{
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s){	br = new BufferedReader(new InputStreamReader(s));}

		public String next() throws IOException 
		{
			while (st == null || !st.hasMoreTokens()) 
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

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

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

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

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

		public boolean ready() throws IOException {return br.ready();} 
	}
}