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 n = s.length;
		int[][] dp = new int[n + 1][n + 1];
		int[] cnt = new int[n + 1];
		for (int len = 1; len <= n; len++) {
			for (int l = 0; (l+len-1) < n; l++) {
				int r = (l + len - 1);
				if (l == r) {
					dp[l][r] = 1;
				} else if (s[l] == s[r]
						&& ((r - l < 2) || (dp[l + 1][r - 1] > 0))) {
					dp[l][r] = Math.min(dp[l][l + (len / 2) - 1], dp[r - (len / 2) + 1][r]) + 1;
				}
				cnt[dp[l][r]]++;
			}
		}
		for (int i = n - 1; i >= 1; i--) {
			cnt[i] += cnt[i + 1];
		}
		for (int i = 1; i <= n; i++) {
			System.out.print(cnt[i] + " ");
		}
		System.out.println();
		sc.close();
	}
}