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 TIM;


import java.io.*;
import java.lang.*;
import java.math.*;
import java.util.*;


public class Main {
    static Scanner input = new Scanner(System.in);
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter pw = new PrintWriter(System.out);

    public static void main(String[] args) throws IOException {
        char[] s = input.next().toCharArray();
        int[][] dp = new int[5005][5005];//dp[i][j]记录区间(i,j)是几阶回文串
        int[] ans = new int[5005];
        for (int len = 1; len <= s.length; len++)//len为当前回文串长度
            for (int l = 0; l <= s.length - len; l++) {
                int r = l + len - 1;
                if (l == r) dp[l][r] = 1;
                else if (s[l] == s[r] && (l + 1 > r - 1 || dp[l + 1][r - 1] > 0))
                    dp[l][r] = Math.min(dp[l][l + (len >> 1) - 1], dp[r - (len >> 1) + 1][r]) + 1;
                ans[dp[l][r]]++;
            }
        for (int i = s.length - 1; i > 0; i--)
            ans[i] += ans[i + 1];
        for (int i = 1; i <= s.length; i++)
            System.out.print(ans[i] + " ");
        System.out.println();
    }
}