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.Arrays;
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
 *
 * @author yittg
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskD solver = new TaskD();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskD {
        int[][] dp = new int[5005][5005];
        int[] ans = new int[5005];

        private int comp(String s, int i, int j) {
            if (dp[i][j] >= 0) {
                return dp[i][j];
            }
            if (s.charAt(i) != s.charAt(j)) {
                dp[i][j] = 0;
                return dp[i][j];
            }
            if (i == j) {
                dp[i][j] = 1;
            } else if (i + 1 == j) {
                dp[i][j] = 2;
            } else if (comp(s, i + 1, j - 1) == 0) {
                dp[i][j] = 0;
            } else {
                dp[i][j] = comp(s, i, (i + j + 1) / 2 - 1) + 1;
            }
            ans[dp[i][j]]++;
            return dp[i][j];
        }

        public void solve(int testNumber, Scanner in, PrintWriter out) {
            String s = in.next();

            for (int[] sub : dp) {
                Arrays.fill(sub, -1);
            }
            for (int i = 0; i < s.length(); ++i) {
                for (int j = i; j < s.length(); ++j) {
                    comp(s, i, j);
                }
            }
            for (int i = 1; i <= s.length(); ++i) {
                int v = 0;
                for (int j = i; j <= s.length(); ++j) {
                    v += ans[j];
                }
                if (i > 1) {
                    out.print(' ');
                }
                out.print(v);
            }
            out.println();
        }

    }

    static class Scanner {
        private BufferedReader br;
        private StringTokenizer st;

        public Scanner(InputStream in) {
            br = new BufferedReader(new InputStreamReader(in));
        }

        public String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

    }
}