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 cfRound427;

import java.util.Scanner;


/**
 * Created with IntelliJ IDEA.
 * Author: yunyansanjin
 * Date: 2017/8/5
 * Time: 2:21
 */
public class D {

    public static void main(String[] args) {
        int[][] dp = new int[5005][5005];
        int[] ans = new int[5005];

        Scanner sc = new Scanner(System.in);
        String s = sc.nextLine();
        int n = s.length();
        for(int i = 0; i < n; i++) dp[i][i] = 1;
        ans[1] = n;
        for(int len = 2; len <= n; len++){
            for(int i = 0; i + len - 1 < n; i++){
                int j = i + len - 1;
                if(s.charAt(i) == s.charAt(j) && (len == 2 || dp[i + 1][j - 1] > 0)){
                    int k = dp[i][(i + j - 1) / 2 ] + 1;
                    dp[i][j] = k;
                    ans[k]++;
                }
            }
        }
        for(int i = n - 1; i > 0; i--){
            ans[i] += ans[i + 1];
        }
        for(int i = 1; i <= n; i++){
            System.out.printf("%d%c", ans[i], i == n ? '\n' : ' ');
        }
    }
}