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;

/**
 * Created by javlon on 09.08.17.
 */
public class Contest {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s = sc.next();
        int n = s.length();
        int[][] dp = new int[n][n];
        int[] ans = new int[n];
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n - i; ++j) {
                if (i == 0) {
                    dp[j][j + i] = 1;
                } else if (i == 1 && s.charAt(j) == s.charAt(j + i)) {
                    dp[j][j + i] = 2;
                } else if (i > 1 && s.charAt(j) == s.charAt(j + i) && dp[j + 1][j + i - 1] != 0) {
                    dp[j][j + i] = dp[j][(2 * j + i + 1) / 2 - 1] + 1;
                } else {
                    continue;
                }
                ans[dp[j][j + i] - 1]++;
            }
        }
        for (int i = n - 2; i >= 0; --i) {
            ans[i] += ans[i + 1];
        }
        for (int i = 0; i < n; ++i) {
            System.out.print(ans[i] + " ");
        }
    }
}