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.StringTokenizer;

public class A {
	static char [] c;
	static int n;
	static final int p = 31, m = (int)1e9+9;
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		String s = sc.next();
		c = s.toCharArray();
		n = c.length;
		compute_hash();
		int [][] res = new int[n][n];
		long [] ans = new long[n+1];
		for (int i = n-1; i >= 0; --i) {
			for (int j = i; j < n; ++j) {
				if((i+1 >= j-1 || res[i+1][j-1] == 1) && c[i] == c[j])
				{
					res[i][j] = 1;
				}
			}
		}
		for (int l = 1; l <= n; ++l) {
			int len = l / 2 - 1;
			for (int i = 0; i+l-1 < n; ++i) {
				int j = i+l-1;
				if(len >= 0 && i+len >= 0 && j-len < n && i+len < j-len && equal(i+1, j+1, len))
				{
					int min = Math.min(res[i][i+len], res[j-len][j]);
					if(min > 0)
						res[i][j] = min+1;
				}
				if(res[i][j]+1 < ans.length)
					--ans[res[i][j]+1];
				++ans[1];
			}
		}
		for (int i = 1; i < ans.length; ++i) {
			ans[i] += ans[i-1];
			out.print(ans[i] + " ");
		}
		out.flush();
		out.close();
	}
	static long [] hash, pow;
	static void compute_hash() {
		hash = new long[n+1];
		pow = new long[n+1];
		hash[0] = 0;
		pow[0] = 1L;
		for (int i = 1; i <= n; ++i){
			hash[i] = (hash[i-1] + (c[i-1] - 'a' + 1) * pow[i-1]) % m;
			pow[i] = (pow[i-1] * p) % m;
		}
	}
	static boolean equal(int i, int j, int len){
		return ((hash[i+len]-hash[i-1] + m)%m*pow[j-len-i] + m)%m == (hash[j] - hash[j-len-1] + m)%m;
	}

	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();
		}
	}
}