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.BufferedWriter;
import java.io.IOException;
import java.io.OutputStreamWriter;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) throws IOException {
        new Solution().run();
    }
}

class Solution {
    void run() throws IOException {
        Scanner in = new Scanner(System.in);
        String s = in.next();
        int n = s.length();
        int[][] dp = new int[n + 1][n + 1];
        int[] cnt = new int[n + 1];
        for (int len = 1; len <= n; len++) {
            for (int l = 0; l + len - 1 < n; l++) {
                int r = l + len - 1;
                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)) {
                    dp[l][r] = Math.min(dp[l][l + len / 2 - 1], dp[r - len / 2 + 1][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] + " ");
        }
        System.out.print("\n");
    }
}