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 round427;
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;

public class D {
	InputStream is;
	PrintWriter out;
	String INPUT = "";
	
	void solve()
	{
		char[] s = ns().toCharArray();
		int n = s.length;
		int[] rs = palindrome(s);
		int[][] h = new int[n][n];
		for(int i = 0;i < n;i++){
			for(int j = 1, k = i, l = i;j <= rs[i*2];j+=2,k--,l++){
				h[k][l] = 1;
			}
		}
		for(int i = 0;i < n-1;i++){
			for(int j = 2, k = i, l = i+1;j <= rs[i*2+1];j+=2,k--,l++){
				h[k][l] = 1;
			}
		}
		for(int i = 2;i <= 14;i++){
			for(int j = 0;j < n;j++){
				for(int k = j+(1<<i-1)-1;k < n;k++){
					int len = k-j+1;
					if(h[j][k] >= i-1 && h[j][j+len/2-1] >= i-1 && h[k-len/2+1][k] >= i-1){
						h[j][k] = i;
					}
				}
			}
		}
		int[] ans = new int[n];
		for(int i = 0;i < n;i++){
			for(int j = i;j < n;j++){
				if(h[i][j] > 0){
					ans[h[i][j]-1]++;
				}
			}
		}
		for(int i = n-2;i >= 0;i--)ans[i] += ans[i+1];
		for(int i = 0;i < n;i++){
			out.print(ans[i] + " ");
		}
		out.println();
	}
	
	public static int[] palindrome(char[] str)
	{
		int n = str.length;
		int[] r = new int[2*n];
		int k = 0;
		for(int i = 0, j = 0;i < 2*n;i += k, j = Math.max(j-k, 0)){
			// normally
			while(i-j >= 0 && i+j+1 < 2*n && str[(i-j)/2] == str[(i+j+1)/2])j++;
			r[i] = j;
			
			// skip based on the theorem
			for(k = 1;i-k >= 0 && r[i]-k >= 0 && r[i-k] != r[i]-k;k++){
				r[i+k] = Math.min(r[i-k], r[i]-k);
			}
		}
		return r;
	}

	
	void run() throws Exception
	{
//		StringBuilder sb = new StringBuilder();
//		int n = 5000;
//		for(int i = 0;i < n;i++){
//			sb.append('a');
//		}
//		INPUT = sb.toString();
		
		is = oj ? System.in : new ByteArrayInputStream(INPUT.getBytes());
		out = new PrintWriter(System.out);
		
		long s = System.currentTimeMillis();
		solve();
		out.flush();
		tr(System.currentTimeMillis()-s+"ms");
	}
	
	public static void main(String[] args) throws Exception { new D().run(); }
	
	private byte[] inbuf = new byte[1024];
	public int lenbuf = 0, ptrbuf = 0;
	
	private int readByte()
	{
		if(lenbuf == -1)throw new InputMismatchException();
		if(ptrbuf >= lenbuf){
			ptrbuf = 0;
			try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
			if(lenbuf <= 0)return -1;
		}
		return inbuf[ptrbuf++];
	}
	
	private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
	private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
	
	private double nd() { return Double.parseDouble(ns()); }
	private char nc() { return (char)skip(); }
	
	private String ns()
	{
		int b = skip();
		StringBuilder sb = new StringBuilder();
		while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ')
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}
	
	private char[] ns(int n)
	{
		char[] buf = new char[n];
		int b = skip(), p = 0;
		while(p < n && !(isSpaceChar(b))){
			buf[p++] = (char)b;
			b = readByte();
		}
		return n == p ? buf : Arrays.copyOf(buf, p);
	}
	
	private char[][] nm(int n, int m)
	{
		char[][] map = new char[n][];
		for(int i = 0;i < n;i++)map[i] = ns(m);
		return map;
	}
	
	private int[] na(int n)
	{
		int[] a = new int[n];
		for(int i = 0;i < n;i++)a[i] = ni();
		return a;
	}
	
	private int ni()
	{
		int num = 0, b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}
		
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	
	private long nl()
	{
		long num = 0;
		int b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}
		
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	
	private boolean oj = System.getProperty("ONLINE_JUDGE") != null;
	private void tr(Object... o) { if(!oj)System.out.println(Arrays.deepToString(o)); }
}