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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayDeque;
import java.util.Arrays;
import java.util.Comparator;
import java.util.Random;
import java.util.StringTokenizer;

public class Solution{
	
	static final int maxn = 5005;
	
    public static void main(String[] args) throws IOException {
     
    	FastScanner fs = new FastScanner();
    	PrintWriter out = new PrintWriter(System.out);
    	
    	
    		
    	int tt = 1;
    	while(tt-->0) {
    		
    		
    		char[] s = fs.next().toCharArray();
    		int n = s.length;
    		
    		int[][] dp = new int[maxn][maxn];
    		int[] cnt = new int[maxn];
    		
    		for(int len=1;len<=n;len++) {
    			for(int l=0;l+len-1<n;l++) {
    				int r = l + len - 1;
    				if(l==r) dp[l][r] = 1;
    				else if(s[l]==s[r] && (l+1>r-1 || dp[l+1][r-1]>0)) {
    					dp[l][r] = dp[l][l+len/2-1] + 1;
    				}
    				cnt[dp[l][r]]++;
    			}
    		}
    		
    		for(int i=n-1;i>=0;i--) {
    			cnt[i] += cnt[i+1];
    		}
    		
    		for(int i=1;i<=n;i++) {
    			out.print(cnt[i]+" ");
    		}
    		out.println();
    		
    		
    		
    		
    		
    		
    		
    	}
    	
    		
    		
    		
    	out.close();
    		
    }


  
    
    
    
    
    
    
    
    
    
    
    
    static final Random random=new Random();
    	
    static void ruffleSort(int[] a) {
    	int n=a.length;//shuffle, then sort 
    	for (int i=0; i<n; i++) {
    		int oi=random.nextInt(n); int temp=a[oi];
    		a[oi]=a[i]; a[i]=temp;
    	}
    	Arrays.sort(a);
    }
   
  
    
    	
    	
    static class FastScanner{
    	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    	StringTokenizer st = new StringTokenizer("");
     
    	public String next(){
    		while(!st.hasMoreElements()){
    			try{
    				st = new StringTokenizer(br.readLine());
    			} catch(IOException e){
    				e.printStackTrace();
    			}
    		}
    		return st.nextToken();
    	}
    		
    	public String nextLine() throws IOException {
    		return br.readLine();
    	}
    		
    	public int nextInt(){
    		return Integer.parseInt(next());
    	}
     
    	public int[] readArray(int n){
    		int[] a = new int[n];
    		for(int i=0;i<n;i++)
    			a[i] = nextInt();
    		return a;
    	}
    		
    	public long nextLong() {
    		return Long.parseLong(next());
    	}
    		
    	public char nextChar() {
    		return next().toCharArray()[0];
    	}
    }
   	
}