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.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskD solver = new TaskD();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskD {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            String s = in.next();
            int[][] dp = new int[s.length()][s.length()];
            for (int i = 0; i < s.length(); i++) {
                dp[i][i] = 1;
            }
            for (int i = 1; i < s.length(); i++) {
                if (s.charAt(i - 1) == s.charAt(i)) {
                    dp[i - 1][i] = 2;
                }
            }
            for (int len = 3; len <= s.length(); len++) {
                for (int first = 0; first < s.length(); first++) {
                    int last = first + len - 1;
                    if (last >= s.length()) break;
                    if (s.charAt(first) == s.charAt(last) && dp[first + 1][last - 1] > 0) {
                        int mid = (first + last - 1) / 2;
                        dp[first][last] = dp[first][mid] + 1;
                    }
                }
            }
            int[] cnt = new int[s.length() + 1];
            for (int first = 0; first < s.length(); first++) {
                for (int last = first; last < s.length(); last++) {
                    cnt[dp[first][last]]++;
                }
            }
            for (int i = s.length(); i > 0; i--) {
                cnt[i - 1] += cnt[i];
            }
            for (int i = 1; i <= s.length(); i++) {
                out.print(cnt[i] + " ");
            }
        }

    }

    static class InputReader {
        private BufferedReader reader;
        private StringTokenizer stt;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream));
        }

        public String nextLine() {
            try {
                return reader.readLine();
            } catch (IOException e) {
                return null;
            }
        }

        public String next() {
            while (stt == null || !stt.hasMoreTokens()) {
                stt = new StringTokenizer(nextLine());
            }
            return stt.nextToken();
        }

    }
}