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.util.*;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        char[] s = sc.next().toCharArray();
        int[] ch = new int[s.length + 1];

        boolean[][] pal = new boolean[s.length][s.length];
        int[][] deg = new int[s.length][s.length];
        for(int i = 0; i < s.length; ++i) {
            deg[i][i] = 1;
            pal[i][i] = true;
        }
        for(int len = 2; len <= s.length; ++len)
            for(int i = 0; i + len <= s.length; ++i) {
                pal[i][i + len - 1] = s[i] == s[i + len - 1] && (len == 2 || pal[i + 1][i + len - 2]);
                deg[i][i + len - 1] = pal[i][i + len - 1] ? deg[i][i + len / 2 - 1] + 1 : 0;
            }
        for(int[] x: deg)
            for(int y: x)
                ch[y]++;
        for(int i = s.length; i > 1; --i)
            ch[i - 1] += ch[i];
        for(int i = 1; i <= s.length; ++i) {
            System.out.print(ch[i] + " ");
        }
    }
}