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.*;
import java.util.*;

public class Codeforces835D {

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(System.out);
		String[] sp = br.readLine().split(" ");
		String word = sp[0];
		int s = word.length();
		char[] charWord = new char[s];
		for (int i = 0; i < s; i++) {
			charWord[i] = word.charAt(i);
		}
		
		//ans[k-1] = number of k-palindromes
		int[] ans = new int[s];
		//only keep track of size 1 palindromes for now
		ans[0] = s;

		//palindrome[i][j] = 1 iff the word starting at char i and ending at char j is a palindrome
		//eventually will become k if the word is a k-palindrome
		int[][] palindrome = new int[s][s];
		for (int i = 0; i < s; i++) {
			palindrome[i][i] = 1;
			int left = i-1;
			int right = i+1;
			boolean isPalindrome = true;
			while ((left >= 0) && (right < s) && (isPalindrome)) {
				if (charWord[left] == charWord[right]) {
					palindrome[left][right] = 1;
					left--;
					right++;
				}
				else {
					isPalindrome = false;
				}
			}
		}
		for (int i = 0; i < s-1; i++) {
			int left = i;
			int right = i+1;
			boolean isPalindrome = true;
			while ((left >= 0) && (right < s) && (isPalindrome)) {
				if (charWord[left] == charWord[right]) {
					palindrome[left][right] = 1;
					left--;
					right++;
				}
				else {
					isPalindrome = false;
				}
			}
		}
		
		for (int diff = 1; diff < s; diff++) {
			for (int i = 0; i < (s-diff); i++) {
				if (palindrome[i][i+diff] > 0) {
					int k = 1 + palindrome[i][i + (diff-1)/2];
					palindrome[i][i+diff] = k;
					ans[k-1]++;
				}
			}
		}
		
		for (int i = s-2; i >= 0; i--) {
			ans[i] += ans[i+1];
		}
		
		for (int i = 0; i < s; i++) {
			System.out.print(ans[i] + " ");
		}
	}
}