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;
        FastScanner in = new FastScanner(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, FastScanner in, PrintWriter out) {
            char[] s = in.next().toCharArray();
            int n = s.length;
            boolean[][] isPal = new boolean[n][n];
            for (int len = 1; len <= n; len++) {
                for (int i = 0; i + len <= n; i++) {
                    int j = i + len - 1;
                    if (len == 1) {
                        isPal[i][j] = true;
                        continue;
                    }
                    if (len == 2) {
                        isPal[i][j] = s[i] == s[j];
                        continue;
                    }
                    isPal[i][j] = s[i] == s[j] && isPal[i + 1][j - 1];
                }
            }
            int[][] d = new int[n][n];
            int[] ans = new int[n + 1];
            for (int len = 1; len <= n; len++) {
                for (int i = 0; i + len <= n; i++) {
                    int j = i + len - 1;
                    d[i][j] = isPal[i][j] ? 1 : 0;
                    if (len == 1) {
                        ++ans[d[i][j]];
                        continue;
                    }
                    if (isPal[i][j]) {
                        d[i][j] = 1 + d[i][i + len / 2 - 1];
                    }
                    ++ans[d[i][j]];
                }
            }
            for (int i = n - 1; i >= 0; i--) {
                ans[i] += ans[i + 1];
            }
            for (int i = 1; i <= n; i++) {
                if (i > 1) {
                    out.print(" ");
                }
                out.print(ans[i]);
            }
            out.println();
        }

    }

    static class FastScanner {
        private BufferedReader in;
        private StringTokenizer st;

        public FastScanner(InputStream stream) {
            in = new BufferedReader(new InputStreamReader(stream));
        }

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

    }
}