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.io.*;
import java.util.*;

public class CF835D {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		char[] aa = br.readLine().toCharArray();
		int n = aa.length;
		int m = n + n - 1;
		char[] bb = new char[m];
		for (int i = 0; i < m; i++)
			bb[i] = i % 2 == 0 ? aa[i / 2] : ' ';
		int[] pq = new int[m];
		pq[0] = 1;
		for (int i = 1, c = 0, r = 1; i < m; i++)
			if (i < r && pq[c + c - i] < r - i)
				pq[i] = pq[c + c - i];
			else {
				c = i;
				if (r < i + 1)
					r = i + 1;
				while (r < m && c + c - r >= 0 && bb[c + c - r] == bb[r])
					r++;
				pq[i] = r - c;
			}
		int[][] cc = new int[n][n];
		int[] dd = new int[n + 1];
		for (int i = 0; i < n; i++)
			cc[i][i] = 1;
		dd[1] = n;
		for (int l = 2; l <= n; l++)
			for (int i = 0, j; (j = i + l - 1) < n; i++) {
				int c = i + j;
				if (pq[c] > j + j - c) {
					int h = c / 2, k;
					if (l % 2 == 0)
						k = Math.min(cc[i][h], cc[h + 1][j]);
					else
						k = Math.min(cc[i][h - 1], cc[h + 1][j]);
					cc[i][j] = k + 1;
					dd[k + 1]++;
				}
			}
		for (int i = n - 1; i >= 1; i--)
			dd[i] += dd[i + 1];
		StringBuilder sb = new StringBuilder();
		for (int i = 1; i <= n; i++)
			sb.append(dd[i] + " ");
		System.out.println(sb);
	}
}