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 D159 {
	public static void main(String[] args){
		Scanner br = new Scanner(System.in);
		String line = br.next();
		BIT b1 = new BIT(line.length()+2);
		BIT b2 = new BIT(line.length()+2);
		for(int i = 0;i<line.length();i++){
			int left = i;
			int right = i;
			while(left >= 0 && right < line.length() && line.charAt(left) == line.charAt(right)){
				b1.inc(right, 1);
				b2.inc(left, 1);
				left--;
				right++;
			}
			left = i;
			right = i+1;
			while(left >= 0 && right < line.length() && line.charAt(left) == line.charAt(right)){
				b1.inc(right, 1);
				b2.inc(left, 1);
				left--;
				right++;
			}
		}
		long ans = 0;
		for(int i = 0;i<line.length()-1;i++){
			ans += b1.sum(i, i) * b2.sum(i+1, line.length()+1);
		}
		System.out.println(ans);
	}
	public static class BIT {
		long[] vals;
		int n; 
		public BIT(int N){
			n = N;
			vals = new long[n];
		}
		// Returns sum of elements with indexes a..b, inclusive
		long sum(int a, int b) {
		    if (a == 0) {
		        long sum = 0;
		        for (; b >= 0; b = (b & (b + 1)) - 1)
		          sum += vals[b];
		        return sum; 
		    } else 
		        return sum(0, b) - sum(0, a-1);
		}
		// Increases value of k-th element by inc.
		void inc(int k, int inc) {
		    for (; k < n; k |= k + 1)
		        vals[k] += inc;
		}
	}
}