Problem: Palindromic characteristics of string s with length |s| is a sequence of |s| integers, where k-th number is the total number of non-empty substrings of s which are k-palindromes.

A string is 1-palindrome if and only if it reads the same backward as forward.

A string is k-palindrome (k > 1) if and only if: 

  1. Its left half equals to its right half. 
  2. Its left and right halfs are non-empty (k - 1)-palindromes. 



The left half of string t is its prefix of length ⌊|t| / 2⌋, and right half — the suffix of the same length. ⌊|t| / 2⌋ denotes the length of string t divided by 2, rounded down.

Note that each substring is counted as many times as it appears in the string. For example, in the string "aaa" the substring "a" appears 3 times.

Input

The first line contains the string s (1 ≤ |s| ≤ 5000) consisting of lowercase English letters.

Output

Print |s| integers — palindromic characteristics of string s.

Examples

Input

abba


Output

6 1 0 0 


Input

abacaba


Output

12 4 1 0 0 0 0 

Note

In the first example 1-palindromes are substring «a», «b», «b», «a», «bb», «abba», the substring «bb» is 2-palindrome. There are no 3- and 4-palindromes here.

Solution (in JAVA):

Solution:
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Scanner;

public class P835D
{
	static char[] s;
	static int[][] isPalin;
	static int[][] maxK;
	public static void main(String[] args)
	{
		Scanner scan = new Scanner(System.in);
		s = scan.next().toCharArray();
		isPalin = new int[s.length][s.length];
		for (int i = 0; i < isPalin.length; i++)
			Arrays.fill(isPalin[i], -1);
		maxK = new int[s.length][s.length];
		for (int i = 0; i < maxK.length; i++)
			Arrays.fill(maxK[i], -1);
		int[] k = new int[20];
		for (int i = 0; i < s.length; i++)
		{
			for (int j = i; j < s.length; j++)
			{
				int m = getMaxK(i, j);
				k[m]++;
			}
		}
		for (int i = 1; i < k.length; i++)
		{
			for (int j = i+1; j < k.length; j++)
			{
				k[i] += k[j];
			}
		}
		PrintWriter pw = new PrintWriter(System.out);
		for (int i = 1; i <= s.length; i++)
		{
			if (i > 1)
				pw.print(" ");
			if (i < k.length)
				pw.print(k[i]);
			else
				pw.print(0);
		}
		pw.flush();
	}
	private static int getMaxK(int a, int b)
	{
		if (maxK[a][b] >= 0)
			return maxK[a][b];
		int ans = -1;
		boolean pal = getIsPalindrome(a, b);
		if (!pal)
			ans = 0;
		else if (a == b)
			ans = 1;
		else
		{
			int mid = (a+b)/2;
			if ((b-a+1) % 2 == 1)
				ans = Math.min(getMaxK(a, mid-1), getMaxK(mid+1, b))+1;
			else
				ans = Math.min(getMaxK(a, mid), getMaxK(mid+1, b))+1;
		}
		return maxK[a][b] = ans;
	}
	private static boolean getIsPalindrome(int a, int b)
	{
		if (b < a)
			return true;
		if (isPalin[a][b] >= 0)
			return isPalin[a][b] == 1;
		boolean ans = false;
		if (a == b)
			ans = true;
		else
			ans = s[a] == s[b] && getIsPalindrome(a+1, b-1);
		isPalin[a][b] = ans ? 1 : 0;
		return ans;
	}
}