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.File;
import java.util.HashSet;
import java.util.Scanner;
import java.util.StringTokenizer;

public class p045 {
	public static void main(String args[]) throws Exception {
// 		StringTokenizer stok = new StringTokenizer(new Scanner(new File("C:/Users/Arunkumar/Downloads/input.txt")).useDelimiter("\\A").next());
		StringTokenizer stok = new StringTokenizer(new Scanner(System.in).useDelimiter("\\A").next());
		StringBuilder sb = new StringBuilder();
		char[] a = stok.nextToken().toCharArray();
		int n = a.length;
		long[] dp = new long[n];
		HashSet<Integer>[] st = new HashSet[n];
		long tot =0;
		for(int i=0;i<n;i++) {
			st[i] = new HashSet<Integer>();
			if(i>0) dp[i]=dp[i-1];
			for(int j=i;j>=0;j--) {
				boolean isp=false;
				if(i==j) {
					st[i].add(j);
					isp=true;
				}
				else if(j==i-1) {
					if(a[j]==a[i]) {
						st[i].add(j);
						isp=true;
					}
				}
				else {
					if(a[j]==a[i] && st[i-1].contains(j+1)) {
						st[i].add(j);
						isp=true;
					}
				}
				if(isp) {
					dp[i]++;
					if(j>0) tot+=dp[j-1];
				}
			}
		}
		System.out.println(tot);
	}
}