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.util.ArrayList;
import java.util.Collections;


public class PalPairs {
	static int b=33;
	static long[] hs;
	static long[] ihs;
	static long[] powB;
	static int n;
	public static void main(String[] args) throws IOException {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		String linea=br.readLine();
		n=linea.length();
		int[] degs=new int[n+1];
		powB=new long[n+1];
		powB[0]=1;
		for (int i=1;i<n+1;i++){
			powB[i]=powB[i-1]*b;
		}
		hs=new long[n+1];
		for (int i=n-1;i>=0;i--){
			hs[i]=hs[i+1]*b+linea.charAt(i);
		}
		ihs=new long[n+1];
		for (int i=n-1;i>=0;i--){
			ihs[i]=ihs[i+1]*b+linea.charAt(n-1-i);
		}
		ArrayList<Integer> ies=new ArrayList<Integer>();
		ArrayList<Integer> jotas=new ArrayList<Integer>();
		for (int i=0;i<n;i++){
			for (int j=i;j<n;j++){
				if (palindrome(i,j)){
					ies.add(i);
					jotas.add(j);
				}
			}
		}
		Collections.sort(ies);
		Collections.sort(jotas);
		long res=0;
		int x=0;
		for (int j:jotas){
			while (x!=ies.size()){
				int i=ies.get(x);
				if (i>j){
					res+=(ies.size()-x);
					break;
				}
				else x++;
			}
		}
		System.out.println(res);
	}
	static boolean palindrome(int i,int j){
		return hash(i,j,hs)==hash(n-1-j,n-1-i,ihs);
	}
	static long hash(int i, int j,long[] hs2) {
		long res=hs2[i]-hs2[j+1]*powB[j+1-i];
		return res;
	}
}