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.util.*;
public class d {
	public static void main(String args[]) {
		Scanner in = new Scanner(System.in);
		String S = in.next();
		
		int[] tree = new int[S.length()];
		ArrayList<Integer> endPoints = new ArrayList<Integer>();
		//Odd length palindromes
		for(int i = 0; i < S.length(); i++) {
			endPoints.add(i);
			increase(tree, i, 1);
			for(int j = 1; j <= i && i+j < S.length(); j++) {
				if(S.charAt(i-j) != S.charAt(i+j))
					break;
				endPoints.add(i+j);
				increase(tree, i-j, 1);
			}
		}
		
		//Even length palindromes
		for(int i = 0; i < S.length(); i++) {
			for(int j = 0; j <= i && j+1+i < S.length(); j++) {
				if(S.charAt(i-j) != S.charAt(i+1+j))
					break;
				endPoints.add(i+1+j);
				increase(tree, i-j, 1);
			}
		}
		
		long count = 0;
		for(int i : endPoints) {
			if(i == S.length())
				continue;
			count += query(tree, i+1, S.length()-1);
		}
		System.out.println(count);
		
	}
	
	static class tuple {
		int s, e;
		public tuple(int s, int e) {
			this.s = s;
			this.e = e;
		}
	}
	
	static int query(int[] tree, int a, int b) {
	    if (a == 0) {
	        int sum = 0;
	        for (; b >= 0; b = (b & (b + 1)) - 1)
	          sum += tree[b];
	        return sum; 
	    } else {
	        return query(tree, 0, b) - query(tree, 0, a-1);
	    }
	}

	
	static void increase(int[] tree, int k, int inc) {
	    for (; k < tree.length; k |= k + 1)
	        tree[k] += inc;
	}
}