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.*;
import java.math.*;
import java.util.concurrent.*;

public final class round_427_d
{
    static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	static FastScanner sc=new FastScanner(br);
    static PrintWriter out=new PrintWriter(System.out);
	static Random rnd=new Random();
	
    public static void main(String args[]) throws Exception
    {
		char[] arr=sc.next().toCharArray();int n=arr.length;
		
		char[] a=new char[n+1];
		
		for(int i=1;i<=n;i++)
		{
			a[i]=arr[i-1];
		}
		
		int[] res=new int[n+1];int[][] val=new int[n+1][n+1];
		
		for(int i=1;i<=n;i++)
		{
			int ptr1=i,ptr2=i;
			
			while(ptr1>=1 && ptr2<=n && a[ptr1]==a[ptr2])
			{	
				val[ptr1][ptr2]++;
				
			//	res[1]++;
				
				ptr1--;ptr2++;
			}
		}
		
		for(int i=1;i<=n;i++)
		{
			int ptr1=i,ptr2=i+1;
			
			while(ptr1>=1 && ptr2<=n && a[ptr1]==a[ptr2])
			{
				val[ptr1][ptr2]++;
				
			//	res[1]++;
				
				ptr1--;ptr2++;
			}
		}
		
	//	out.println(Arrays.toString(res));
		
		int[][] high=new int[n+1][n+1];
		
		for(int i=1;i<=n;i++)
		{
			for(int j=i;j<=n;j++)
			{
				if(i==j)
				{
					high[i][j]=1;
					
					res[high[i][j]]++;
					
					continue;
				}
				
				
				if(val[i][j]==1)
				{
					int len=(j-i+1);
					
					high[i][j]=high[i][i+(len/2)-1]+1;
					
					res[high[i][j]]++;
				}
			}
		}
		
		int sum=0;
		
	//	out.println(Arrays.toString(res));
		
		for(int i=n;i>=1;i--)
		{
			sum+=res[i];
			
			res[i]=sum;
		}
		
		for(int i=1;i<=n;i++)
		{
			out.print(res[i]+" ");
		}
		
		out.println("");out.close();
    }
}
class FastScanner
{
    BufferedReader in;
    StringTokenizer st;

    public FastScanner(BufferedReader in) {
        this.in = in;
    }
	
    public String nextToken() throws Exception {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }
	
	public String next() throws Exception {
		return nextToken().toString();
	}
	
    public int nextInt() throws Exception {
        return Integer.parseInt(nextToken());
    }

    public long nextLong() throws Exception {
        return Long.parseLong(nextToken());
    }

    public double nextDouble() throws Exception {
        return Double.parseDouble(nextToken());
    }
}