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.util.Scanner;

public class D {
	public static void main(String []args) {
		Scanner cin = new Scanner(System.in);
		int[][] dp = new int[5005][5005];
		int[] ans = new int[5005];
		String s = cin.nextLine();
		int n = s.length();
		for(int d = 1;d <= n;d ++)
			for(int i = 0;i + d - 1< n;i ++) {
				int j = i + d - 1;
				if(d < 3) dp[i][j] = (s.charAt(i) == s.charAt(j) ? d : 0);
				else if(s.charAt(i) != s.charAt(j) || dp[i + 1][j - 1] == 0) dp[i][j] = 0;
				else if(s.charAt(i) == s.charAt((i + j - 1) / 2)) dp[i][j] = dp[i][(i + j - 1) / 2] + 1;
				else dp[i][j] = 1;
				ans[dp[i][j]] ++;
			}
		for(int i = n - 1;i > 0;i --)
			ans[i] += ans[i + 1];
		for(int i = 1;i <= n;i ++)
			System.out.printf("%d ", ans[i]);
	}
}