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;

public class ProblemD {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);

        char[] s = in.nextLine().toCharArray();
        int n = s.length;

        int[][] DP = new int[n][n];

        for (int i = 0; i < n; i++)
            DP[i][i] = 1;

        for (int i = 1; i < n; i++) {
            if (s[i-1] == s[i])
                DP[i-1][i] = 2;
        }

        for (int t = 2; t < n; t++) {
            for (int l = 0, r = t; r < n; l++, r++) {
                if (s[l] == s[r] && DP[l+1][r-1] > 0) {
                    int m = (l + r + 1) / 2 - 1;
                    DP[l][r] = DP[l][m] + 1;
                }
            }
        }

        int[] count = new int[n + 1];
        for (int i = 0; i < n; i++)
            for (int j = i; j < n; j++)
                count[DP[i][j]]++;

        for (int i = n - 1; i >= 0; i--)
            count[i] += count[i+1];

        for (int i = 1; i <= n; i++)
            System.out.print(count[i] + " ");
        System.out.println();
    }

}