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.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Abhas Jain
 */
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 {
        int[][] dp = new int[5001][5001];

        public void solve(int testNumber, InputReader in, PrintWriter out) {
            String x = in.next();
            int n = x.length();
            int[] count = new int[n + 1];
            for (int i = 0; i < n; ++i) {
                dp[i][i] = 1;
            }
            for (int i = 0; i < n - 1; ++i) {
                if (x.charAt(i) == x.charAt(i + 1)) dp[i][i + 1] = 2;
            }
            for (int j = 2; j < n; ++j) {
                inner:
                for (int i = 0; i + j < n; ++i) {
                    if (x.charAt(i) != x.charAt(i + j) || dp[i + 1][i + j - 1] == 0) continue inner;
                    int curr = (i + i + j) / 2;
                    if (j % 2 == 0) curr--;
                    dp[i][i + j] = dp[i][curr] + 1;
                }
            }
            for (int i = 0; i < n; ++i) {
                for (int j = 0; j < n; ++j) {
                    count[dp[i][j]]++;
                }
            }
            int[] prefix = new int[n + 1];
            prefix[n] = count[n];
            for (int i = n - 1; i > -1; --i) {
                prefix[i] = count[i] + prefix[i + 1];
            }
            for (int i = 1; i <= n; ++i) out.print(prefix[i] + " ");
            // out.print("0");
        }

    }

    static class InputReader {
        private InputStream stream;
        private byte[] buf = new byte[1024];
        private int curChar;
        private int numChars;
        private InputReader.SpaceCharFilter filter;

        public InputReader(InputStream stream) {
            this.stream = stream;
        }

        public int read() {
            if (numChars == -1) {
                throw new InputMismatchException();
            }
            if (curChar >= numChars) {
                curChar = 0;
                try {
                    numChars = stream.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (numChars <= 0) {
                    return -1;
                }
            }
            return buf[curChar++];
        }

        public String nextString() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            StringBuilder res = new StringBuilder();
            do {
                if (Character.isValidCodePoint(c)) {
                    res.appendCodePoint(c);
                }
                c = read();
            } while (!isSpaceChar(c));
            return res.toString();
        }

        public boolean isSpaceChar(int c) {
            if (filter != null) {
                return filter.isSpaceChar(c);
            }
            return isWhitespace(c);
        }

        public static boolean isWhitespace(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

        public String next() {
            return nextString();
        }

        public interface SpaceCharFilter {
            public boolean isSpaceChar(int ch);

        }

    }
}