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.IOException;
import java.util.Scanner;

public class Bakery {
	
	
	// author : qoo2p5
	

	public static void main(String[] args) throws IOException {
			
	//	BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		
		Scanner in = new Scanner(System.in);
		
		char[] word = in.nextLine().toCharArray();
		
		int n = word.length;
		
		int[][] dp = new int[n][n];
		long[] arr = new long[n+1];

		for(int i=0; i<n; i++) {
			
			for(int j=0; j<n; j++) {
					
				
				int left = j;
				int right = j + i;
				
				if(left < 0 || right >= n) {
					continue;
				}
				
				if(left == right) {
					dp[left][right] = 1;
					arr[dp[left][right]]++;
					continue;
				}
				
				if(left + 1 == right) {
					if(word[left] == word[right]) {
						dp[left][right] = 2;
						arr[dp[left][right]]++;
					}
					
					continue;
				}
				
				if(word[left] != word[right] || dp[left+1][right-1] == 0) {
					continue;
				}
				
				int mid = (left + right + 1)/2 - 1;
				
				dp[left][right] = dp[left][mid] + 1;
				
				arr[dp[left][right]]++;
				
				
			
				
			}
		}
		
		
		for(int i=n-2; i>=1; i--) {
			arr[i] += arr[i+1];
		}
		
		for(int i=1; i<=n; i++) {
			System.out.print(arr[i] + " ");
		}
		
		
	}	
}