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 Class_1 {
 
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		String s = in.next();
		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.charAt(l)==s.charAt(r)&&(l+1>r-1||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 j=n-1;j>=1;j--){
			cnt[j]+=cnt[j+1];
		}
		for(int i=1;i<=n;i++){
			System.out.print(cnt[i]+" ");
		}
		System.out.println("\n");
	}
 
}