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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Comparator;
import java.util.Iterator;
import java.util.PriorityQueue;
import java.util.StringTokenizer;
import java.util.TreeSet;
import java.io.BufferedReader;
import java.io.InputStreamReader;

/**
 * Built using CHelper plug-in Actual solution is at the top
 */
public class Main {
	public final static double MAX_T = 1000000;
	public final static double MIN_T = 0;
	public static int s = 0;

	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskC solver = new TaskC();
		solver.solve(1, in, out);
		out.close();
	}

	static class TaskC {
		class poli {
			int x;
			int length;
		}

		public void solve(int testNumber, InputReader in, PrintWriter out) {
			char[] s = in.next().toCharArray();
			int n = s.length;
			char[][] dp = new char[n][n];
			boolean[][] eq = new boolean[n][n];
			for (int i = 0; i < n; i++) {
				eq[i][i] = true;
				for (int j = i + 1; j < n; j++) {
					eq[i][j] = s[i] == s[j];
				}
			}
			for (int i = 0; i < n; i++) {
				dp[i][i] = 1;
			}
			for (int i = 1; i < n; i++) {
				if (eq[i - 1][i]) {
					dp[i - 1][i] = 2;
				}
			}
			for (int d = 2; d < n; d++) {
				for (int i = 0, j = d, j2 = i + ((d + 1) >> 1) - 1; j < n; i++, j++, j2++) {
					if (eq[i][j] && dp[i + 1][j - 1] != 0) {
						dp[i][j] = (char) (dp[i][j2] + 1);
					}
				}
			}
			int[] ans = new int[n + 1];
			for (int i = 0; i < n; i++) {
				for (int j = 0; j < n; j++) {
					ans[dp[i][j]]++;
				}
			}
			for (int i = n - 1; i > 0; i--) {
				ans[i] += ans[i + 1];
			}
			for (int i = 1; i <= n; i++) {
				out.print(ans[i] + " ");
			}
			out.println();
		}

	}

	static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream), 32768);
			tokenizer = null;
		}

		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

		public long nextLong() {
			return Long.parseLong(next());
		}

		public double nextFloat() {
			return Double.parseDouble(next());
		}

		public double nextDouble() {
			return Double.parseDouble(next());
		}

	}
}