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:
//package e_contest_11;

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

public class palindromePairs {
		static int [][] memo;
		static char[] c;
		static int dp(int start, int end) {
			if(start ==end)
				return memo[start][end] =1;
			if(start==end-1) {
				return memo[start][end] = c[start]==c[end]?2:0;
			}
//			if(start>end)
//				return 0;
			if(memo[start][end]!=-1)
				return memo[start][end];
			if(c[start]!=c[end])
				return memo[start][end] = 0;
			int pal = dp(start+1, end-1);
			if(pal==0)
				return memo[start][end] = 0;
			int mid = (start+end)/2;
			if((start+end)%2==0)
				mid--;
//			if(start==1&& end==2)
//				System.out.println(start+" "+end+" "+mid);
			return memo[start][end] = 1 + dp(start,mid);
		}
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String s = sc.next();
		c = s.toCharArray();
		int n = c.length;
		memo = new int[n][n];
		for (int i = 0; i < memo.length; i++) {
			Arrays.fill(memo[i], -1);
		}
		for (int i = 0; i < memo.length; i++) {
			for (int j = i; j < memo.length; j++) {
				dp(i,j);
//				System.out.println(memo[i][j]+" "+s.substring(i,j+1));
			}
		}
		int k = (int)Math.ceil(Math.log(n)/Math.log(2));
		int [] cnt = new int[k+1];
		for (int i = 0; i < memo.length; i++) {
			for (int j = 0; j < memo.length; j++) {
				if(memo[i][j]!=-1)
					cnt[memo[i][j]]++;
			}
		}
		long ansK[]= new long[k+1];
		long cntSofar = 0;
		for (int i = cnt.length-1; i >=1; i--) {
			cntSofar+=cnt[i];
			ansK[i] = cntSofar;
		}
		PrintWriter out =new PrintWriter(System.out);
		for (int i = 1; i < ansK.length; i++) {
			out.print(ansK[i]+" ");
		}
		for (int j = ansK.length-1; j < n; j++) {
			out.print("0 ");
		}
		out.flush();
	}
}