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.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        String s = in.next();
        int n = s.length();

        int[][] maxPalin = new int[n + 1][n + 1];
        int[] count = new int[n + 1];

        for (int subLen = 1; subLen <= n; subLen++) {
            for (int left = 0; left <= n - subLen; left++) {
                int right = left + subLen - 1;

                if (subLen == 1) {
                    maxPalin[left][left] = 1;
                }
                else if (subLen == 2) {
                    if (s.charAt(left) == s.charAt(right)) {
                        maxPalin[left][right] = 2;
                    }
                    else {
                        maxPalin[left][right] = 0;
                    }
                }
                else {
                    if (s.charAt(left) == s.charAt(right) && maxPalin[left + 1][right - 1] > 0) {
                        maxPalin[left][right] = maxPalin[left][left + subLen / 2 - 1] + 1;
                    }
                }

                count[maxPalin[left][right]]++;
            }
        }

        for (int i = n - 1; i >= 1; i--) {
            count[i] += count[i + 1];
        }

        for (int i = 1; i <= n; i++) {
            System.out.print(count[i]);
            System.out.print(' ');
        }

        System.out.print('\n');
    }
}