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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main2 {
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        char[] arr = reader.readLine().toCharArray();
        int n = arr.length;
        int[][] dp = new int[n][n];
        for (int i=0; i<n; i++) {
            int l = i, r = i;
            while (l>-1 && r<n) {
                if (arr[l]!=arr[r]) break;
                dp[l][r] = 1;
                l--; r++;
            }
            l = i-1; r = i;
            while (l>-1 && r<n) {
                if (arr[l]!=arr[r]) break;
                dp[l][r] = 1;
                l--; r++;
            }
        }
        for (int i=0; i<n; i++) {
            for (int j=i+1; j<n; j++) {
                if (dp[i][j]>0 && dp[i][i + ((j-i+1)/2) - 1]>0) {
                    dp[i][j] = Math.max(dp[i][j], dp[i][i + ((j-i+1)/2) - 1] + 1);
                }
            }
        }
        int[] result = new int[n+1];
        for (int i=0; i<n; i++) {
            for (int j=0; j<n; j++) {
                result[dp[i][j]]++;
            }
        }
        for (int i=n-1; i>0; i--) {
            result[i]+=result[i+1];
        }
        StringBuilder ans = new StringBuilder();
        for (int i=1; i<=n; i++) ans.append(result[i]).append(" ");
        System.out.println(ans);
    }
}