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 PalindromicCharacteristics {
	//dp[l][r] is the maximum k such that the substring built from characters from l to r is k-palindrome.
	static int[][] dp;
	static StringBuilder sb;
	public static void main(String[] args) {
		//get input
		Scanner s = new Scanner(System.in);
		char[] str = s.nextLine().toCharArray();
		s.close();
		//set starting variables
		dp = new int[str.length][str.length];
		sb = new StringBuilder();
		//fill in dp
		//trivial values
		for (int l = 0; l < dp.length; l++) {
					dp[l][l] = 1;
		}
		int[] count = new int[str.length+1];
		count[1] += dp.length;
		//dynamic values
		for (int len = 2; len <= str.length; len++) {
			for (int l = 0; l <= str.length-len; l++) {
				int r = l + len - 1;
				if(str[r]==str[l] && (l + 1 > r - 1 || dp[l+1][r-1]!=0)) {
					dp[l][r] = 1+Math.min(dp[l][l+len/2-1],dp[r-len/2+1][r]);
				}
				for (int i = 1; i <= dp[l][r]; i++) {
					count[i]++;	
				}
			}
		}
		for (int i = 1; i <= str.length; i++) {
			sb.append(count[i] + " ");
		}
		System.out.println(sb);
	}
}