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()

dp = [[0]*5005 for _ in range(5005)]
n = len(s)
ans = [0 for _ in range(5005)]

for length in range(1,n+1):
	for l in range(n-length+1):
		r = l+length
		if(length == 1):
			dp[l][r] = 1
			continue
		elif(length == 2):
			dp[l][r] = 2 if(s[l] == s[r-1]) else 0
			continue
		if(s[l] != s[r-1] or dp[l+1][r-1] == 0):
			continue
		dp[l][r] = 1
		m = (l+r) // 2
		if(length&1):
			if(dp[l][m] and dp[m+1][r]):
				dp[l][r] = dp[l][m]+1
		else:
			if(dp[l][m] and dp[m][r]):
				dp[l][r] = dp[l][m]+1

for length in range(1,n+1):
	for l in range(n-length+1):
		ans[dp[l][l+length]] += 1

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

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

print()