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:
//package com.company;

import java.io.*;
import java.util.*;

public class Main {
    public static void main(String[] args) throws IOException {
        MyScanner sc = new MyScanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        PalindromicCharacteristics.solve(sc, pw);
        pw.close();
    }
    //http://codeforces.com/contest/835/problem/D
    public static class PalindromicCharacteristics {

        public static void solve(MyScanner sc, PrintWriter pw) throws IOException {
            char[] s = sc.next().toCharArray();
            int n = s.length;
            int[][] dp = new int[n + 1][n + 1];
            for (int i = 0; i < n; i++) {
                dp[i][i+1] = 1;
            }
            for (int i = 0; i < n - 1; i++) {
                if (s[i] == s[i + 1]) {
                    dp[i][i + 2] = 2;
                } else {
                    dp[i][i + 2] = 0;
                }
            }
            for (int i = 3; i <= n; i++) {
                for (int j = 0; j + i <= n; j++) {
                    if (s[j + i - 1] != s[j] || dp[j + 1][j + i - 1] == 0) {
                        dp[j][i + j] = 0;
                        continue;
                    }
                    dp[j][i+j] = dp[j][(i + j + j) / 2] + 1;
                }
            }
            int[] cnt = new int[n + 1];
            for (int i = 0; i < n; i++) {
                for (int j = i + 1; j <= n; j++) {
                    cnt[dp[i][j]]++;
                }
            }
            int su = 0;
            for (int i = n; i >= 0; i--) {
                cnt[i] += su;
                su = cnt[i];
            }
            for (int i = 1; i <= n; i++) {
                pw.print(cnt[i] + " ");
            }
            pw.println();
        }




    }



    static class MyScanner {
        StringTokenizer st;
        BufferedReader br;

        public MyScanner(InputStream s){  br = new BufferedReader(new InputStreamReader(s));}

        public MyScanner(FileReader s) throws FileNotFoundException {br = new BufferedReader(s);}

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

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

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

        public String nextLine() throws IOException {return br.readLine();}

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

        public boolean ready() throws IOException {return br.ready();}
    }
}