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 PYTHON3):

Solution:
s = input()
n = len(s)
max_palin = [[0 for i in range(n + 1)] for j in range(n + 1)]
count = [0 for i in range(n + 1)]

for sub_len in range(1, n + 1):
    for left in range(0, n - sub_len + 1):
        right = left + sub_len - 1

        if sub_len == 1:
            max_palin[left][right] = 1
        elif sub_len == 2:
            if s[left] == s[right]:
                max_palin[left][right] = 2
            else:
                max_palin[left][right] = 0
        else:
            if s[left] == s[right] and max_palin[left + 1][right - 1] > 0:
                max_palin[left][right] = max_palin[left][left + sub_len // 2 - 1] + 1

        count[max_palin[left][right]] += 1

for i in range(n - 1, 0, -1):
    count[i] += count[i + 1]

for i in range(1, n + 1):
    print(count[i], end=' ')

print()