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

        int n = s.length();
        int [][] arr = new int[n][n];
        int [] res = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i][i] = 1;
            res[0]++;
            if (i > 0) {
                if (s.charAt(i-1) == s.charAt(i)) {
                    arr[i-1][i] = 2;
                    res[1]++;
                }
            }
        }
        for (int len = 3; len <= n; len++) {
            for (int i = 0; i < n-len+1; i++) {
                int j = i + len-1;
                if (s.charAt(i) == s.charAt(j) && arr[i+1][j-1] > 0) {
                    int l = i-1 + len/2, r = j+1 - len/2;
                    int k = Math.min(arr[i][l], arr[r][j]) + 1;
                    arr[i][j] = k;
                    res[k-1]++;
                }
            }
        }
        for (int i = n-2; i >= 0; i--) {
            res[i] += res[i+1];
        }
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < n; i++) {
            if (i > 0) sb.append(" ");
            sb.append(res[i]);
        }
        System.out.println(sb.toString());
    }
}