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().strip()
n = len(s)
pl = [0]*(n+1)

dp = [[0]*n for i in range(n)]

for i in range(n):
	for j in range(n-i):
		l = j+i
		if i == 0:
			dp[j][l] = 1
			pl[1] +=1
		elif  i == 1:
			if s[j] == s[l]:
				dp[j][l] = 2
				pl[2] += 1
		else:
			if s[j] == s[l] and dp[j+1][l-1] > 0:
				if (i+1)%2 == 0:
					mid = int((i+1)/2)-1
					dp[j][l] = dp[j][j+mid]  + 1
					pl[dp[j][l]] += 1
				else:
					mid = int((i+1)/2)-1
					dp[j][l] = dp[j][j+mid]  +1	
					pl[dp[j][l]] += 1

ans = ''
for i in range(n-1,-1,-1):
	pl[i] += pl[i+1]
for i in  range(1,n+1):
	ans = ans + str(pl[i]) + ' '
print(ans)