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.*;
import java.util.Arrays;
import java.util.StringTokenizer;


public class Main {
    static final int UNCALC = -1;
    static char[] s;
    static int memo[][];

    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        s = sc.next().toCharArray();
        int n = s.length;
        memo = new int[n][n];
        for (int len = 1; len <= n; len++) {
            for (int l = 0; l < n && l+len-1 < n; l++) {
                int r = l + len - 1;
                if (l == r) memo[l][r] = 1;
                else if (s[l] == s[r] && (memo[l + 1][r - 1] > 0|| l+1 > r-1))
                    memo[l][r] = Math.min(memo[l][l + len / 2 - 1], memo[r - len / 2 + 1][r])+1;
            }
        }
        int ans[] = new int[n + 1];
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++) {
                ans[memo[i][j]]++;
            }
//        System.err.println(Arrays.toString(ans));
        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.flush();
        out.close();
    }


    static class Scanner {
        BufferedReader br;
        StringTokenizer st;

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

        String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

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

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

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


}