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.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
public class B {
	
	public static void main(String[] args) throws IOException  {
		Scanner sc = new Scanner(System.in);
		PrintWriter wr = new PrintWriter(System.out);
		String s=sc.next();
		int n=s.length();
		int dp[][]=new int [n][n];
		int cnt[]=new int [n+1];
		for(int i=0;i<n;i++)dp[i][i]=1;
		for(int i=1;i<=n;i++) {
			for(int l=0;l+i-1<n;l++) {
				int r=l+i-1;
				//System.out.println(l+1+" "+(r-1)+"****"+l+" "+r);
				if(l==r)dp[l][l]=1;
				else if(i==2)dp[l][r]=s.charAt(l)==s.charAt(r)?2:0;
				else if(s.charAt(l)!=s.charAt(r)||(dp[l+1][r-1]==0))dp[l][r]=0;
				else {
					dp[l][r]=dp[l][(int) Math.ceil((l+r)/2.0)-1]+1;
				}
				cnt[dp[l][r]]++;

			}
		}
		int ans[]=new int[n+1];
		ans[n]=cnt[n];
		//System.out.println(Arrays.toString(cnt));

		for(int i=n-1;i>0;i--) {
			ans[i]+=ans[i+1]+cnt[i];
		}
		//System.out.println(Arrays.toString(ans));
		for(int i=1;i<=n;i++)System.out.print(ans[i]+" ");
		
		
		wr.close();
	}
}


class Scanner 
{
	StringTokenizer st;
	BufferedReader br;

	public Scanner(InputStream s){	br = new BufferedReader(new InputStreamReader(s));}

	public String next() throws IOException 
	{
		while (st == null || !st.hasMoreTokens()) 
			st = new StringTokenizer(br.readLine());
		return st.nextToken();
	}

	public int nextInt() throws IOException {return Integer.parseInt(next());}
	
	public long nextLong() throws IOException {return Long.parseLong(next());}

	public String nextLine() throws IOException {return br.readLine();}
	
	public double nextDouble() throws IOException
	{
		String x = next();
		StringBuilder sb = new StringBuilder("0");
		double res = 0, f = 1;
		boolean dec = false, neg = false;
		int start = 0;
		if(x.charAt(0) == '-')
		{
			neg = true;
			start++;
		}
		for(int i = start; i < x.length(); i++)
			if(x.charAt(i) == '.')
			{
				res = Long.parseLong(sb.toString());
				sb = new StringBuilder("0");
				dec = true;
			}
			else
			{
				sb.append(x.charAt(i));
				if(dec)
					f *= 10;
			}
		res += Long.parseLong(sb.toString()) / f;
		return res * (neg?-1:1);
	}
	
	public boolean ready() throws IOException {return br.ready();}


}