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.*;
import java.util.*;

public class PalindromePairs {
	public static void main(String args[]) throws Exception{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String s= br.readLine();
		int n = s.length();
		boolean pal[][]= new boolean[n][n];
		for(int i=0;i<n;i++){
			pal[i][i] = true;
		}
		
		for(int l=2;l<=n;l++){
			for(int i=0;i<n-l+1;i++){
				int j = i+l-1;
				if(l==2 && s.charAt(i)== s.charAt(j)){
					pal[i][j] = true;
				}
				else if(s.charAt(i)== s.charAt(j) && pal[i+1][j-1])
					pal[i][j] = true;
			}
		}
		
		long ans=0;
		
		long dp[][] = new long[n][n];
		for(int l=1;l<=n;l++){
			for(int i=0;i<n-l+1;i++){
				int j = i+l-1;
				if(l==1){
					dp[i][j] =1;
				}
				else if(l==2){
					dp[i][j] = dp[i][j-1] + dp[i+1][j];
					if(s.charAt(i) == s.charAt(j))
						dp[i][j]++;
				}
				else{
					dp[i][j] = dp[i][j-1] + dp[i+1][j] - dp[i+1][j-1];
					if(s.charAt(i) == s.charAt(j) && pal[i+1][j-1])
						dp[i][j]++;
				}
			}
		}
		
		for(int l=1;l<=n;l++){
			for(int i=0;i<n-l+1;i++){
				int j = i+l-1;
				
				if(pal[i][j]){
					if(j+1<n)
						ans += dp[j+1][n-1];
				}
			}
		}
		System.out.println(ans);
	}
}