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 vkd{
	public static void main(String[] args){
		Scanner br = new Scanner(System.in);
		String line = br.next();
		ArrayList<Integer> endps = new ArrayList<Integer>();
		BIT sps = new BIT(line.length()+1);
		for(int i = 0;i<line.length();i++){
			int leftp = i;
			int rightp = i;
			while(leftp >= 0 && rightp < line.length() && line.charAt(leftp) == line.charAt(rightp)){
				endps.add(rightp);
				sps.inc(leftp+1, 1);
				leftp--;
				rightp++;
			}
		}
		for(int i = 0;i<line.length()-1;i++){
			int leftp = i;
			int rightp = i+1;
			while(leftp >= 0 && rightp < line.length() && line.charAt(leftp) == line.charAt(rightp)){
				endps.add(rightp);
				sps.inc(leftp+1, 1);
				leftp--;
				rightp++;
			}
		}
		long ans = 0;
		for(int i = 0;i<endps.size();i++){
			ans = ans + sps.sum(endps.get(i)+2, line.length());
		}
		System.out.println(ans);
	}
	public static class BIT {
		int n;
		int[] v;
		BIT(int N) {
			n=N;
			v = new int[n];
		}
		int sum(int lo, int hi) {
			if(lo==0) {
				int sum = 0;
				for(int i = hi;i>=0; i=(i&(i+1))-1)
					sum += v[i];
				return sum;
			}
			return sum(0,hi)-sum(0,lo-1);
		}
		void inc(int p, int x) {
			for(int i=p; i<n; i|=(i+1))
				v[i]+=x;
		}
	}
}