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)
isPalindrome = [[False for i in range(n)] for i in range(n)]
for i in range(n):
    isPalindrome[i][i] = True
for i in range(n - 1, -1, -1):
    for j in range(i + 1, n):
        if (s[i] == s[j] and (i + 1 == j or isPalindrome[i + 1][j - 1] == True)):
            isPalindrome[i][j] = True

degreePalindrome = [[0 for i in range(n)] for i in range(n)]
#degreePalindrome[i][j] = degreePalindrome[i][mid] + 1
res = [0] * (n + 1)
for i in range(n):
    for j in range(i, n):
        if (i == j):
            degreePalindrome[i][j] = 1
        elif isPalindrome[i][j]:
            mid = (i + j - 1) // 2
            degreePalindrome[i][j] = degreePalindrome[i][mid] + 1
        res[degreePalindrome[i][j]] += 1

for i in range( n - 2, 0, -1):
    res[i] += res[i + 1]

print(*res[1::])