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.IOException;
import java.io.PrintWriter;
import java.net.Inet4Address;
import java.util.*;
import java.io.*;

public class Main {
    public void solve() throws IOException {
        String s = nextToken();
        int n = s.length();
        int[] res = new int[n + 1];
        int[][] a = new int[n][n];
        for (int i = 0; i < n; i++) {
            a[i][i] = 1;
            res[a[i][i]]++;
        }
        for (int i = 1; i < n; i++) {
            for (int j = 0; j + i < n; j++) {
                if ((a[j + 1][j + i - 1] > 0 || i == 1) && s.charAt(j) == s.charAt(j + i)) {
                    a[j][j + i] = 1;
                    a[j][j + i] = Math.min(a[j][j + (i + 1) / 2 - 1], a[j + (i + 1) / 2  + ((i + 1) % 2)][j + i]) + 1;
                    res[a[j][j + i]]++;
                }
            }
        }
        int t = 0;
        for(int i = n; i >= 0; i--){
            t += res[i];
            res[i] = t;
        }
        for(int i = 1; i <= n; i++){
            out.print(res[i] + " ");
        }
    }


    BufferedReader br;
    StringTokenizer sc;
    PrintWriter out;

    public String nextToken() throws IOException {
        while (sc == null || !sc.hasMoreTokens()) {
            try {
                sc = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                return null;
            }
        }
        return sc.nextToken();
    }

    public boolean hasNext() throws IOException {
        while (sc == null || !sc.hasMoreTokens()) {
            try {
                sc = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                return false;
            }
        }
        return true;
    }

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

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

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

    public static void main(String[] args) throws IOException {
        try {
            Locale.setDefault(Locale.US);
        } catch (Exception e) {
        }
        new Main().run();
    }

    public void run() {
        try {
            br = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
//            br = new BufferedReader(new FileReader("a"));
//            out = new PrintWriter(new File("a"));
            solve();
            out.close();
        } catch (Exception e) {
            e.printStackTrace();
            System.exit(1);
        }
    }
}