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

public class ProblemD {

	static Scanner sc = new Scanner(System.in);
	static PrintWriter pw = new PrintWriter(System.out);

	public static void main(String[] args) throws IOException {
		String s = sc.next();

		int n = s.length(), dp[][] = new int[n + 1][n + 1];
		int cnt[] = new int[n + 1];

		for (int k = 1; k <= n; k++) { // loop on all possible pal lengths
			for (int l = 0; l + k - 1 < n; l++) {
				int r = l + k - 1; // end index of the substring start at l with
									// length k
				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)) {
					int mid1 = l + k / 2 - 1, mid2 = r - k / 2 + 1;
					dp[l][r] = Math.min(dp[l][mid1], dp[mid2][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] + " ");
		}
		pw.flush();
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

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

		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 double nextDouble() throws IOException {
			String x = next();
			StringBuilder sb = new StringBuilder("0");
			double res = 0, f = 1;
			boolean dec = false, neg = false;
			int start = 0;
			if (x.charAt(0) == '-') {
				neg = true;
				start++;
			}
			for (int i = start; i < x.length(); i++)
				if (x.charAt(i) == '.') {
					res = Long.parseLong(sb.toString());
					sb = new StringBuilder("0");
					dec = true;
				} else {
					sb.append(x.charAt(i));
					if (dec)
						f *= 10;
				}
			res += Long.parseLong(sb.toString()) / f;
			return res * (neg ? -1 : 1);
		}

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

		public int[] nextIntArray(int n) throws IOException {
			int[] array = new int[n];
			for (int i = 0; i < n; i++) {
				array[i] = nextInt();
			}
			return array;
		}
	}
}