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:
//package CF;

import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Stack;
import java.util.StringTokenizer;

public class A {

	public static void main(String[] args) throws Exception {
		Scanner bf = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		char[] s = bf.next().toCharArray();
		int n = s.length;
		boolean[][] palin = new boolean[n][n];
		int[][] memo = new int[n][n];
		for (int i = 0; i < palin.length; i++) {
			palin[i][i] = true;
			memo[i][i] = 1;
			if (i + 1 < n)
				palin[i][i + 1] = s[i] == s[i + 1];
		}
		for (int i = n - 1; i >= 0; i--) {
			for (int j = 2; j + i < palin.length; j++) {
				palin[i][j + i] = palin[i + 1][j + i - 1] && s[i] == s[j + i];
			}
		}
		int[] ans = new int[n];
		for (int i = n - 1; i >= 0; i--) {
			for (int j = 1; j + i < palin.length; j++) {
				if (palin[i][i + j])
					memo[i][i + j] = 1 + Math.min(memo[i][i + (j-1) / 2], memo[i+j - (j-1) / 2][j + i]);
			}
		}
		for (int i = 0; i < memo.length; i++) {
//			System.out.println(Arrays.toString(memo[i]));
			for (int j = 0; j < memo.length; j++) {
				if (memo[i][j] > 0)
					ans[memo[i][j] - 1]++;
			}
		}
		for (int i = n-2; i >= 0; i--) {
			ans[i] += ans[i+1];
		}
		for (int i : ans)
			out.print(i + " ");
		out.println();
		out.flush();
		out.close();
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public Scanner(FileReader fileReader) {
			br = new BufferedReader(fileReader);
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public boolean ready() throws IOException {
			return br.ready();
		}
	}
}