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;
import java.util.ArrayList;

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

    private final int NAXN = 5001;

    private Scanner in;
    private BufferedWriter out;
    private String s;
    private int n;
    private int dp[][];
    private int cnt[];

    private void run() throws IOException {
        in = new Scanner(System.in);
        out = new BufferedWriter(new OutputStreamWriter(System.out));
        
        s = in.next();
        n = s.length();
        dp = new int[n][n];
        cnt = new int[n + 1];
        int k = 1;

        for (int len = 1; len <= n; ++len) {
            for (int i = 0; i < n - len + 1; ++i) {
                int j = i + len - 1;
                dp[i][j] = 0;
                if (len == 1) {
                    if (s.charAt(i) == s.charAt(j)) {
                        dp[i][j] = 1;
                    }
                } else if (len == 2) {
                    if (s.charAt(i) == s.charAt(j)) {
                        dp[i][j] = 2;
                    }
                } else {
                    int m = len / 2;
                    if (s.charAt(i) == s.charAt(j) && dp[i + 1][j - 1] > 0) {
                        dp[i][j] = dp[i][i + m - 1] + 1;
                    }
                }
                cnt[dp[i][j]]++;
            }
        }
        
        for (int i = n - 1; i > 0; --i) {
            cnt[i] += cnt[i + 1];
        }

        for (int i = 1; i <= n; ++i) {
            out.write(cnt[i] + " ");
        }

        out.flush();
    }
}